Prove that $(\frac{\partial z}{\partial x})^{2}-(\frac{\partial z}{\partial y})^{2}=\frac{\partial z}{\partial a}\frac{\partial z}{\partial b}$ 
Given $x=a+b,y=a-b$, prove that $$\left(\frac{\partial z}{\partial x}\right)^{2}-\left(\frac{\partial z}{\partial y}\right)^{2}=\frac{\partial z}{\partial a}\frac{\partial z}{\partial b}$$

$z=f(x,y)$
It is the first time that I've seen this weird greek sign.
is it something like $dx/dy$ ? what is going on in this question?
I know that is $d/dy$ is more like an operator than a fraction. Or perhaps I'm just mixing up 2 different things..? sorry it just that I'm not familiar with this particular sign...
 A: Hint.
$$
\frac{\partial x}{\partial a}+\frac{\partial y}{\partial a} = 2\\
\frac{\partial x}{\partial b}-\frac{\partial y}{\partial b} = 2\\
\frac{\partial x}{\partial a}-\frac{\partial y}{\partial a} = 0\\
\frac{\partial x}{\partial b}+\frac{\partial y}{\partial b} = 0\\
$$
and
$$
\frac{\partial z}{\partial x} = \frac{\partial z}{\partial a}\frac{\partial a}{\partial x}+\frac{\partial z}{\partial b}\frac{\partial b}{\partial x}\\
\frac{\partial z}{\partial y} = \frac{\partial z}{\partial a}\frac{\partial a}{\partial y}+\frac{\partial z}{\partial b}\frac{\partial b}{\partial y}
$$
then
$$
\left(\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}\right)\left(\frac{\partial z}{\partial x}-\frac{\partial z}{\partial y}\right)=\cdots
$$
A: $x=a+b,\quad y=a-b\implies a=\frac{1}{2}(x+y)\quad \text{and}\quad b=\frac{1}{2}(x-y)$
By chain rule $$\frac{\partial z}{\partial x} = \frac{\partial z}{\partial a}\frac{\partial a}{\partial x}+\frac{\partial z}{\partial b}\frac{\partial b}{\partial x}$$ and $$\frac{\partial z}{\partial y} = \frac{\partial z}{\partial a}\frac{\partial a}{\partial y}+\frac{\partial z}{\partial b}\frac{\partial b}{\partial y}$$
So $$\frac{\partial z}{\partial x} = \frac{\partial z}{\partial a}\frac{1}{2}+\frac{\partial z}{\partial b}\frac{1}{2}=\frac{1}{2}(\frac{\partial z}{\partial a}+\frac{\partial z}{\partial b})$$ and $$\frac{\partial z}{\partial y} = \frac{\partial z}{\partial a}\frac{1}{2}+\frac{\partial z}{\partial b}(-\frac{1}{2})=\frac{1}{2}(\frac{\partial z}{\partial a}-\frac{\partial z}{\partial b})$$
Therefore $$\left(\frac{\partial z}{\partial x}\right)^{2}-\left(\frac{\partial z}{\partial y}\right)^{2}=\frac{1}{4}\{(\frac{\partial z}{\partial a}+\frac{\partial z}{\partial b})^2-(\frac{\partial z}{\partial a}-\frac{\partial z}{\partial b})^2\}=\frac{1}{4}(4\frac{\partial z}{\partial a}\frac{\partial z}{\partial b})=\frac{\partial z}{\partial a} \frac{\partial z}{\partial b}$$
Hence $$\left(\frac{\partial z}{\partial x}\right)^{2}-\left(\frac{\partial z}{\partial y}\right)^{2}=\frac{\partial z}{\partial a} \frac{\partial z}{\partial b}$$
