Why can we say that the graphs of $f(x,y)=x^2+y^2$ and $g(x,y)=-x^2-y^2+xy^3$ "are tangent to each other" at $(0,0)$ ?
I have done the following:
The tangent plane at the graph of $f(x,y)=x^2+y^2$ at $(0,0)$ is:
$$z=f(0,0)+\left[ \frac{\partial{f}}{\partial{x}}(0,0) \right](x-0)+\left[ \frac{\partial{f}}{\partial{y}}(0,0) \right](y-0)$$
The partial derivatives are:
$$\frac{\partial{f}}{\partial{x}}=2x$$
and
$$\frac{\partial{f}}{\partial{y}}=2y$$
We have that: $f(0,0)=0 \\ \frac{\partial{f}}{\partial{x}}(0,0)=0 \\ \frac{\partial{f}}{\partial{y}}(0,0)=0$
So the equation of the plane that is tangent at the graph of $f$ at $(0,0)$ is $z=0$.
How can we conclude that the graphs of $f(x,y)=x^2+y^2$ and $g(x,y)=-x^2-y^2+xy^3$ "are tangent to each other" at $(0,0)$ ???
Two surfaces are tangent at the point $P\in\mathbb{R}^3$ iff they are both containing $P$ and sharing the same tangent plane at $P$.
equivalently:
The graphs of $f,g\in\mathbb{R}^2\rightarrow \mathbb{R}$ are tangent to each other at $P\in\mathbb{R}^2$ iff they are sharing the same tangent plane at $P$, i.e. $f(P)+\nabla f(P)(X-P) = g(P)+\nabla g(P)(X-P)$, for all $X\in\mathbb{R}^2$
Now, just find the equation of the plane that is tangent at the graph of $g$ at $(0,0)$, exactly as you did for $f$, make sure it's $z=0$, and if so- they are tangent at the origin.