The change of variables $x=uv$, $y=\frac{1}{2}\left(u^2-v^2\right)$ transforms $f(x,y)$ to $g(u,v).$
If $\left\|\nabla f(x,y)\right\|^2=2$ for all $x$ and $y$, determine constants $a$ and $b$ such that
$$a\left(\frac{\partial g}{\partial u}\right)^2-b\left(\frac{\partial g}{\partial v}\right)^2=u^2+v^2.$$
Since $\frac{\partial g}{\partial u}=\frac{\partial f}{\partial x}v+\frac{\partial f}{\partial y}u$ and $\frac{\partial g}{\partial v}=\frac{\partial f}{\partial x}u-\frac{\partial f}{\partial y}v$
$a\left(\frac{\partial g}{\partial u}\right)^2-b\left(\frac{\partial g}{\partial v}\right)^2=\frac{\partial f}{\partial x}^2 \left(av^2-bu^2\right)+\frac{\partial f}{\partial y}^2\left(au^2-bv^2\right)+2\frac{\partial f}{\partial x}\frac{\partial f}{\partial y}uv(a+b)$
I cannot find a way to manipulate this to use the condition $\left\|\nabla f(x,y)\right\|^2=2$ and find the values for $a$ and $b$.
I would greatly appreciate any help.
Note that $ \left\|\nabla f \right\| ^2 = \left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2 = 2$.
Picking up on where you left, lets write down a system of equations:
$$ \begin{cases} \left(\dfrac{\partial f}{\partial x}\right)^2 + \left(\dfrac{\partial f}{\partial y}\right)^2 = 2 \\ \dfrac{\partial f}{\partial x}^2 \left(av^2-bu^2\right)+\dfrac{\partial f}{\partial y}^2\left(au^2-bv^2\right)+2\dfrac{\partial f}{\partial x}\dfrac{\partial f}{\partial y}uv(a+b) = u^2 + v^2 \end{cases} $$
Collecting coefficients in front of the different powers of $u$ and $v$ in the second equation, and assuming non-trivial case $u\not\equiv0$ and $v\not\equiv0$, we write
$$ \begin{cases} f_x^2 + f_y^2 = 2 \\ f_x^2\left(av^2-bu^2\right)+ f_y^2\left(au^2-bv^2\right)+ 2\,f_x \,f_y\, uv(a+b) = u^2 + v^2 \end{cases} \iff \\ \iff \begin{cases} f_x^2 + f_y^2 = 2 \\ u^2\left(a \,f_y^2 - b\,f_x^2 - 1 \right) + v^2\left(a \,f_x^2 - b\,f_y^2 - 1 \right) + uv\left( \,f_x \, f_y \,\left( a+b\right)\right) = 0 \end{cases} \implies \\ \implies \begin{cases} f_x^2 + f_y^2 = 2 \\ a \,f_y^2 - b\,f_x^2 = 1\\ a \,f_x^2 - b\,f_y^2 = 1 \\ a+b = 0 \end{cases} $$ In this way you have a system of for equations for the four unknowns, which are $a,b,f_x$, and $f_y$. Let us sum the second and the third equation:
$$ \begin{cases} f_x^2 + f_y^2 = 2 \\ a \,f_y^2 - b\,f_x^2 = 1\\ a \,f_x^2 - b\,f_y^2 = 1 \\ a+b = 0 \end{cases} \implies \begin{cases} f_x^2 + f_y^2 = 2 \\ a \left( f_x^2 + f_y^2\right) - b\left( f_x^2 + f_y^2\right) = 2\\ a+b = 0 \end{cases} \implies \begin{cases} a - b = 1\\ a+b = 0 \end{cases} \implies \begin{cases} a = \frac{1}{2}\\ b = -\frac{1}{2} \end{cases} $$