Math question functions help me? I have to find find $f(x,y)$ that satisfies 
\begin{align}
f(x+y,x-y) &= xy + y^2 \\
f(x+y, \frac{y}x ) &= x^2 - y^2
\end{align}
So I first though about replacing $x+y=X$ and $x-y=Y$ in the first one but then what?
 A: Hint:
If $x+y=X$ and $x-y=Y$ then 
\begin{align*}
x&=X-y \implies Y= X-y-y \implies 2y=X-Y \dots
\end{align*}
And be careful with the second for $x=0$.
A: $$f(x,y)=f(\frac{x+y}{2}+\frac{x-y}{2},\frac{x+y}{2}-\frac{x-y}{2})$$
$$f(x,y)=f(\frac{x}{1+y}+\frac{xy}{1+y},\frac{\frac{xy}{1+y}}{\frac{x}{1+y}})$$
Edit: Be careful about the domain in the second case. In the end you are looking on the transformation $\mathbb R^2\to\mathbb R^2$ which maps $(x,y)\mapsto (x+y,y/x)$ and want to find an inverse. This won't work, since the function is neither one-to-one nor onto and not even defined everywhere. A priori you know nothing about the function $f$ for points $(0,y)$ with $y\neq -1$ and $(x,-1)$ with $x\neq 0$. However if you write down an explicit formula for $f$ you will find that you can extend $f$ naturally to all points $(x,y)$ with $y\neq -1$.
A: Then express $x,y$ in terms of the new $X,Y$, and you're just there..
Similarly for the second question, $X:=x+y,\ \ Y:=y/x$, then $X/x=Y+1$, which leads to $x$ in terms of $X,Y$...
