Let $f(x, y)$ differentiable function $\forall (x,y)$. Let $g(u, v) = f(u^2 - v^2, u^2v).$ Known $\nabla f(-3, 2) = (2, 1).$ Find $ \nabla g(1, 2)$ I need an explanation for this solution.
I tried to solved the following exercise:

Let $f(x, y)$ differentiable function $\forall (x,y)$. Let $g(u, v) = f(u^2 - v^2, u^2v).$ Known $\nabla f(-3, 2) = (2, 1).$ Find  $ \nabla g(1, 2)$

But with no success.
I looked at the solution and the solution is: $\nabla g(1, 2) = (8, -7).$
Can you please explain me why? Thanks in advance.
 A: The expresion $g(u,v)=f(u^2-v^2,u^2v)$
must be viewed as a composition
$g=f\circ\phi$ where $\phi:{\Bbb{R}}^2\to{\Bbb{R}}^2$
given by
 $$\left(\begin{array}{c}
u\\
v
\end{array}\right)
\mapsto
\left(\begin{array}{c}
u^2-v^2\\
u^2v
\end{array}\right),
$$
is a change of coordinates.
So by the chain's rule we have
$$\nabla g=\nabla(f\circ\phi)=\nabla fJ\phi.$$
Or, evaluated at a position $p$, as
$$\nabla g|_p=\nabla(f\circ\phi)|_p=\nabla f|_{\phi(p)}J\phi|_p.$$
But 
$J\phi=
\left(\begin{array}{cc}
2u&-2v\\
2uv&u^2
\end{array}\right)$.
So
$$\nabla g(u,v)=\nabla f(u,v)\left(\begin{array}{cc}
2u&-2v\\
2uv&u^2
\end{array}\right)
.$$
And, on the use of your data
$$\nabla g(1,2)=\nabla f(-3,2)\left(\begin{array}{cc}
2u&-2v\\
2uv&u^2
\end{array}\right)|_{(1,2)},
$$
which is
$$\nabla g(1,2)=(2,1)\left(\begin{array}{cc}
2&-4\\
4&1
\end{array}\right)=[8,-7].
$$
A: when u = 1, v = 2, u^2 - v^2 = -3 and v*u^2 = 2. This gives: gradient g (1,2) = gradient f (-3,2) * A. Where A is the matrix:
2u      -2v
2uv       u^2
Putting u = 1, v = 2 into A we have: A = 
2      -4
4       1
So gradient of g (1,2) = (2,1)*A = (8, -7)
