@PedroTamaroff @robjohn @DanielFischer @MikeMiller @BalarkaSen

$$f(x,y)=-4x^2y^2+(x^2+y^2)^3$$
$$g(x,y)=-y^3+3x^2y+(x^2+y^2)^2$$

$$P=(0,0)$$


$$f(x,y)=-4x^2y^2+(x^2+y^2)^3</math>: <math>f_4(x,y)=-4x^2y^2 \Rightarrow m_P(f)=4$$

$$g(x,y)=-y^3+3x^2y+(x^2+y^2)^2: g_3(x,y)=-y^3+3x^2y \Rightarrow m_P(g)=3$$

The tangents in $P=(0,0)$
of $f$ are $x=0$ and $y=0$
of $g$ are $y(3x^2-y^2)=0 \Rightarrow y=0, \sqrt{3}x+y=0, \sqrt{3}x-y=0$

So, $f$ and $g$ have a coomon tangent, $y=0$.

So, the intersection isn't tranversal.