Determine $f(x, y)$ if $f(x + y, x - y) = xy + y^2$. 
Determine $f(x, y)$ if $f(x + y, x - y) = xy + y^2$.

I have been doing similar problems, but can't get my head around this one. I'd rather get a tip instead of the full answer. Thanks to anyone who contributes.
My Solution After Tips:
Let $u = x + y$ and $ v = x - y$. Then $ x = u - y $ and $ y = x - v $. From these equations, we have that
$ x = u - y = u - (x - v) \rightarrow x = u - x + v \rightarrow 2x = u + v \Rightarrow x = (u + v)/2 $.
Similarly, $u - v = (x + y) - (x - y) = x + y - x + y = 2y \Rightarrow y = (u - v)/2$.
Substituting, we get
$$ f(u, v) = (\frac{u + v}{2})(\frac{u - v}{2}) + (\frac{u + v}{2})^2 $$

$$ f(u, v) = \frac{u^2 - uv}{2} $$


You can solve it directly as follows:


*

*$f(x+y,x-y) = xy+y^2 = (x+y)y$
Now, write 


*

*$y = \frac{1}{2}((x+y)-(x-y))$
It follows:
$$f(x+y,x-y)= (x+y)\cdot \frac{1}{2}((x+y)-(x-y)) = \frac{1}{2}\left( (x+y)^2 - (x+y)(x-y)\right)$$ $$\Rightarrow f(x,y) = \frac{1}{2}(x^2-xy)$$
