If you mean "touches the surface at an angle" then please ask it that way. To most people, "crossing the surface" would mean that you are on one side and then the other, for example like the way the function $y = x^3$ is on one side of $y=0$ and then the other.

It occurs to me that we never established the point in the exposition of multivariable calculus that you have reached, how you got there, and what context you're in. You'll get a different level of rigor from different books, from applied math versus pure math, from physics vs. applied math, and so forth. So perhaps they skipped the part where they defined what a derivative actually is.

Even for one dimension, the intuition of the epsilon-delta proof may be a little hard to come by. For continuity, we have a rectangular box centered on (x_0, f(x_0)) of width 2*delta and height 2*epsilon, and as we squeeze the height of the box down to zero we're able to simultaneously manipulate the width of the box so that no part of the function "pokes out" through the top or bottom of the box.

But for a derivative, it's not just the value of the function we're concerned with, it's the slopes of the secant lines from (x_0,f(x_0)) to all the nearby points on the curve. The epsilon in the proof is a restriction on how different the secant slope can be from the tangent slope. We're able to make this difference as small as we want by adjusting delta, which in this case limits how far left or right on the curve we allow the secant to hit.

For a function like f(x)=x^2, no secant line has the same slope as the tangent line. Every secant intersects the tangent at an angle. But by reducing delta and forcing the secants to come from more nearby points, you can reduce that angle. We conclude that when the curve of the function finally touches the tangent line, it must do so at zero angle since there's no other angle it can come from.

For two variables it's somewhat similar, except now delta doesn't control the distance between two vertical lines on either side of x_0; instead it controls the radius of a cylinder around the point (x_0,y_0) with axis perpendicular to the x,y plane. And if we look at the part of the function within that cylinder, we have a plane which we're trying to prove is the tangent plane, containing lines through (x_0,y_0,f(x_0,y_0)) that we're trying to prove are tangent lines.

Also within the cylinder we have a part of the graph of the function, and we're interested in the secant lines between (x_0,y_0,f(x_0,y_0)) and all other points of the function in that region. Now epsilon is a restriction on the difference of the slope of the secant line from the slope of the would-be tangent line directly above or below it. And by adjusting delta we can make that difference less than epsilon for every point in the region.

Or rather, we have to prove we can do that for whichever function we're looking at. And then we know the function is differentiable.

When I said we don't have to do epsilon-delta all the time in single-variable calculus, I meant that if we are doing a rigorous course we DO have to do some epsilon-delta proofs at the beginning. But then we do a proof that the product of two differentiable functions is differentiable, and suddenly we don't have to do a separate epsilon-delta proof for x^2 and x^3 and x^4; we just prove that f(x)=x is differentiable and then multiply the function by itself and apply the product rule.

We prove an addition rule, and now we have that all polynomials are differentiable. We prove the chain rule, and now we can compose differentiable functions and know the composition is differentiable. We see a lot less of the epsilon-delta after that and a lot more just finding the answers.

@DavidK thank you!!! Yes you're right. And I think you also predicted correctly that I have no formal introduction to multi-variable calculus yet, just references here and there from a physics book. However, I now understand your answer a LOT better after watching some lectures on YouTube!

Look at my update to the question, and let me know what you think!

I think you've gotten the idea. You probably understand it better now than the vast majority of people who are exposed to partial derivatives, because they don't ask as many questions.

And there's actually some hard work to put in on some functions to prove that they don't actually stick out at angles from the would-be "tangent" plane. Fortunately for physics, the functions they use are mostly all composed of simpler functions that were proved differentiable many years ago, and they ways they are put together are also known to produce differentiable functions. Because I am sure very few physicists have the patience to go back to epsilon-delta proofs.

Got you! That's pretty cool though, that there IS a way to prove that functions are differentiable...I mean, at least to me who hasn't had much exposure to it, it seems a little magical that we could somehow show that it doesn't stick out at angles in ANY direction. Also, although I understand it a lot better now, it still seems like there would be a way for the partial derivatives to (as an example) be zero in the $x$ and $y$ directions, but the surface curves a little bit upwards and....

...and then downwards as we rotate around, although its still differentiable.

Actually, wait a sec...no, the more I think about it, that does seem impossible. If the derivatives in the $x$ and $y$ directions are zero, and the slope is not $0$ in any combination of those directions...then it MUST have a sharp bend...I think...bleh, I just wish I could visualize it a bit better.

I guess that'll come with time. Anyways, thanks a lot David! I'll probably be back with more questions when I get into formal epsilon-delta proofs!