Show that $4(u^2+v^2)\left(\left(\frac{\partial z}{\partial x} \right)^2+\left(\frac{\partial z}{\partial y} \right)^2\right)$ Let $x = u^2-v^2$ and $y = 2uv$, and suppose that $z = f(x,y)$ is differentiable. Show that:
$$ \left(\frac{\partial z}{\partial u}\right) ^2+\left(\frac{\partial z}{\partial v}
\right)^2 =  4(u^2+v^2)\left(\left(\frac{\partial z}{\partial x} \right)^2+\left(\frac{\partial z}{\partial y} \right)^2\right)$$
By taking the derivative in respect to $u$ and $v$ then taking their composite values:
$\frac{d}{dx}=2u-2v; \frac{d}{dy} = 2v+2u$
Which are equivalent to: $2(u-v)2(u+v)=4(u^2-v^2)$ However the signs are different as I'm expected to get a positive between $u$ and $v$.
 A: Sketch for the solution: let $\varphi (u,v):=(u^2-v^2,2uv)=(x,y)$, then you want to show that
$$
\Delta (f\circ \varphi )(u,v)=4(u^2+v^2)\Delta f(x,y)
$$
where $\Delta $ is the Laplacian, defined for a differentiable function $g:\mathbb{R}^n\to \mathbb{R}$ by
$$
\Delta g(p)=\sum_{k=1}^n (\partial_k)^2g(p)=\sum_{k=1}^n\partial ^2g(p)[e_k,e_k]
$$
where $\partial g(p)$ is the Fréchet derivative of $g$ at $p$, the $e_1,\ldots ,e_n$ is the standard orthonormal basis of $\mathbb{R}^n$ and $\partial _k$ is the $k$-th partial derivative. Now, the chain rule and the product rule gives
$$
\begin{align*}
(\partial _k)^2 (f\circ \varphi )(p)&=\partial ^2(f\circ \varphi )(p)[e_k,e_k]\\&=\partial (\partial (f\circ \varphi)(p)e_k)e_k\\&=\partial ((\partial f\circ \varphi )(p)\partial \varphi (p)e_k)e_k\\
&=\partial ((\partial f\circ \varphi )(p)\partial _k\varphi (p))e_k\\&=\partial ((\nabla f\circ \varphi) (p)\cdot \partial _k\varphi (p))e_k\\
&=\partial(\nabla f\circ \varphi )(p)e_k\cdot \partial _k\varphi (p)+\nabla f(\varphi (p))\cdot (\partial _k)^2\varphi (p)\\
&=(\partial\nabla f\circ \varphi )(p)\partial _k \varphi (p)\cdot \partial _k\varphi (p)+\nabla f(\varphi (p))\cdot (\partial _k)^2\varphi (p)\\
&=H_f (\varphi(p))\partial _k \varphi (p)\cdot \partial _k\varphi (p)+\nabla f(\varphi (p))\cdot (\partial _k)^2\varphi (p)\\
&=\partial _k \varphi (p)^\top H_f (\varphi(p))\partial _k\varphi (p)+\nabla f(\varphi (p))\cdot (\partial _k)^2\varphi (p)
\end{align*}
$$
where $\nabla f(q)$ and $H_f(q)$ are the gradient and the Hessian of $f$ at $q$ respectively. Adding up you get
$$
\Delta (f\circ \varphi )(p)=\nabla f(\varphi (p))\cdot \Delta \varphi (p)+\partial _1 \varphi (p)^\top H_f (\varphi(p))\partial _1\varphi (p)+\partial _2 \varphi (p)^\top H_f (\varphi(p))\partial _2\varphi (p)
$$
Now just simplify the above knowing that $[\partial \varphi (u,v)]=\left[\begin{smallmatrix}\partial _1 \varphi (u,v)&\partial _2\varphi (u,v)\end{smallmatrix}\right]=2\left[\begin{smallmatrix}u&-v\\v&u\end{smallmatrix}\right]$ and $\Delta \varphi (u,v)=0$.
