Chain Rule of Partial Derivatives If $f$ is a differentiable function defined $f: \mathbb{R}^2 \to \mathbb{R}$ and $f(2,1) = 3$ and $\nabla f(2,1) = (4,3)$, find $\nabla G(1,1)$ with $G(x,y) := x^2yf(x^2 +y^2, xy)$.
I wrote the $G_x$ as $2xyf(x^2 +y^2, xy) + x^2yf'(x^2 + y^2, xy)f_x(x^2 + y^2, xy)$ but I don't know what the value of $f'(x^2 + y^2, xy)$ is.
Thanks in advance

Denote
$$g(x,y)=x^{2}y$$
Then by product rule
$$\nabla G=f\nabla g+g\nabla f$$
$$\nabla g=2xy\boldsymbol{i}+x^{2}\boldsymbol{j}$$
Now write
$$\begin{cases}
x^{2}+y^{2} & =2\\
xy & =1
\end{cases}$$
Multiply the second equation by 2, add and subtract from the first one obtaining respectively
$$\left(x+y\right)^{2}=4$$
$$\left(x-y\right)^{2}=0$$
Hence
$$x=y=\pm1$$
Finally
$$\nabla G\left(1,1\right)=3\left(2\boldsymbol{i}+1\boldsymbol{j}\right)+4\boldsymbol{i}+3\boldsymbol{j}=10\boldsymbol{i}+6\boldsymbol{j}$$
