# Partial Derivative Chain Rule on a Vector Valued Function

I found this question on a past exam, but I am having a hard time figuring it out since the material was only briefly mentioned in class, without any examples, and is not present in the class notes.

Let $f:\mathbb{R}^2\rightarrow\mathbb{R}$ be twice continuously partially differentiable and let $g:\mathbb{R}^2\rightarrow\mathbb{R}^2$ be defined by $g(x,y)=(x^2-2y^2,xy^2)$. Let $F(x,y)=(f\circ g)(x,y)$. Use the chain rule to express $\frac{\partial^2F}{\partial x\partial y}(x,y)$ in terms of the partial derivatives of $f$.

I'm sure I'll understand the chain rule for vector valued functions once I see an example like this.

$F(x,y)$ = ($f($$x^2$-2$y^2$), $f$(x$y^2$)). This now boils down to a problem using the single variable calculus chain rule. So, $\partial{F} \over\partial{y}$ $=$ ($df \over dy$(${f}$($x^2 - 2y^2$)($-4y$), $df \over dy$($f$($xy^2$)($2xy$)). Next, $\partial^2{F} \over \partial{x} \partial{y}$ $=$ ($df \over dx$($df \over dy$($f$($x^2 - 2y^2)$(-4$y$)),($df \over dx$($df\over dy$($f$($xy^2$)$)($2xy$)$ + ($df \over dy$(f($xy^2)$)($2y$)))