How to find the maximum of this function $\frac{x}{x-y}$? I have a function 
$$f(x,y)=\dfrac{x}{x-y},\;\text{for all}\;x\neq y$$
Where $f(x,y)$ is maximum?
I have calculated $\partial f/\partial x = -y/(x-y)^2$ and $\partial f/\partial y = 1/(x-y)^2$ but what to do next?
 A: Consider
$$g(y) = f(1, y) = \frac1{1-y}$$
Even if $y\neq x = 1$, we can consider the following limit:
$$\lim_{y\to 1^-}\frac1{1 - y} = \infty$$
It should be clear that no maximum exists
A: If your domain is $\mathbb{R}^{2}$, then $f$ doesn't have a maximum. You can very easily make $f$ arbitrarily large by choosing $y$ close to $x$ (provided $x \ne 0$). Since $\frac{\partial f}{\partial y}\ne 0$, the only place $f$ could possible have a maximum is on its boundary (eg if $|y-x|>k$ for some $k$ and all $x,y$ in your domain)
A: Remember that $\frac{\partial f}{\partial x}$ is the slope at a point $(x, y)$ strictly considering the x direction only. Similarly, $\frac{\partial f}{\partial y}$ is the slope at a point strictly considering the y direction only.
For us to have a maximum, both slopes on both directions at that maximum should be $0$. So we get a system:
$$0=\frac{-y}{(x - y)^2}, 0 = \frac{1}{(x - y)^2}$$
Clearly, there is no solution for that second equation, so there is no point in which $\frac{\partial f}{\partial y}$ is zero. That means we don't get a flat surface anywhere in the y-direction, so no maximum exists.
