pde: $x \frac{\partial f}{\partial x} = y \frac{\partial f}{\partial y}$ Bring the partial differential equation into a simpler form:
$x \frac{\partial f}{\partial x} = y \frac{\partial f}{\partial y}$
by making use the fact that the gradiant of f is a "covariant vector" of a "barred coordinate system":
$\bar{x} = xy$
$\bar{y} = (y)^2$
$f(x,y) = f\bigg(x(\bar{x}),~ y(\bar{y}) \bigg) = F(\bar{x}, \bar{y})$. 
says the jacobian is:
$J = \begin{bmatrix}y & x \\ 0 & 2y\end{bmatrix}$
and inverse jacobian is:
$J^{-1} = \begin{bmatrix} \frac{1}{y} & \frac{-x}{2y^2} \\ 0 & \frac{1}{2y}\end{bmatrix} = \bar{J}(x)$
anyways... it turns out that in the barred coordinate system $\bar{y}$ is independent of function F, which means: $F(\bar{x}, \bar{y}) = F(\bar{x})$. 
and general solution is:  $f = F(\bar{x} = xy)$ or simply: $f = F(xy)$
i'm not really sure what that means...F(xy)... that could be anything with one parameter... how does that simplify the pde?
(From Schaum's Outlines, Tensor Calculus, 2011, page 3.7, exercise 3.7)

ok.. i'll try one more time:
pde given by exercise:
$x \frac{\partial f}{\partial x} = y \frac{\partial f}{\partial y}$
The first part of the exercise proves that f = F(xy).
Then, we can substitute this expression into the pde to obtain:
$x \frac{\partial F(xy)}{\partial x} = y \frac{\partial F(xy)}{\partial y}$
now let u(x,y) = xy
$x \frac{\partial F(u)}{\partial x} = y \frac{\partial F(u)}{\partial y}$
apply chain rule:
$x \frac{\partial }{\partial u}[F(u)] \frac{d}{dx}[u(x,y)]= y \frac{\partial }{\partial u}[F(u)] \frac{\partial}{\partial y}[u(x,y)]$
now cancel $\frac{\partial }{\partial u}[F(u)]$ on each side of equation, we have:
$x  \frac{\partial}{\partial x}[u(x,y)]= y  \frac{\partial}{\partial y}[u(x,y)]$
resubsituting $u(x,y) = xy$, we have:
$x  \frac{\partial}{\partial x}[xy]= y  \frac{\partial}{\partial y}[xy]$
$xy  \frac{\partial }{\partial x}[x]= xy  \frac{\partial}{\partial y}[y]$
$\frac{\partial}{\partial x}[x]= \frac{\partial}{\partial y}[y]$
thus simplifying pde.
1 = 1???
why does it simplify down to 1 = 1?
 A: How does that simplify the pde? The PDE has infinitely many solutions. As you can check $f(x,y) = xy$, $f(x,y) = 1 + (xy)^2$ and $f(x,y) = e^{\pi xy}$ are all solutions. The result you have found is the most general form any solution takes. It's as much as there is possible to say about the solution without adding more restrictions.
However it already tells you a lot about how a solution behaves. For example from $f(x,y) = F(xy)$ it follows directly that any solution $f$ takes the same value along all lines $xy = C$. So if you are given any solution and you move along the curve $y = C/x$ and try to evaluate $f(x,y)$ you will find that the function value stays the same (insights like this is the basis for a more general method for solving PDEs known as the method of characteristics).
Why do we have infinitely many solutions? It's because the PDE is not fully defined. A general PDE needs to have some boundary and/or initial conditions to be uniquely defined. This is the same as when we say that $y' = y$ has the solution $y = Ae^x$ for any constant $A$ and that we need to specify, say, $y(0)$ in order to fix $A$ and have a unique solution. If you are to solve the PDE given an initial condition like $f(x,1) = g(x)$ for some given function $g(x)$ (let's say $g(x) = e^x$) then from $f(x,y)=F(xy)$ you automatically get $f(x,y) = g(xy)$ (so $f(x,y) = e^{xy}$ for our simple example). Note that from just knowing the fact that the solution has this particular form $F(xy)$ the solution for all $x,y$ follows directly from the initial condition. This shows that the information you have derived about the solution is indeed very useful.
A: I am very new to PDE's but i would tackle the problem like this:
Assume $f(x,y)=X(x)Y(y).$
Therefore
$xX'(x)Y(y)=X(x)Y'(y)y \Rightarrow$ $$X(x)/(xX{'}(x))=Y(y)/(Y{'}(y)y)=p\in \mathbb{R}$$ $\Rightarrow X_p(x)=xpX_p{'}(x)$ and $Y_p(y)=ypY_p'(y)\Rightarrow$
$$X_p(x)=\sqrt[p]{x}$$
$$Y_p(y)=\sqrt[p]{y}$$
then we get
$f_p(x,y)=\sqrt[p]{xy}$ is a solution for all $p\in\mathbb{R}$. Therefore all linear combinations are also solutions and we get $$f(x,y)=\sum_{p\in\mathbb{R}}c_p\sqrt[p]{xy},\text{ with } c_p\in\mathbb{R}$$ which can be expessed as a function $F(xy)$.
A: Since
$$
(x,-y)\cdot\nabla f=0\tag1
$$
and the Chain Rule says
$$
\frac{\mathrm{d}f}{\mathrm{d}t}=\frac{\partial f}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}t}+\frac{\partial f}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}t}\tag2
$$
we have that $f$ is constant along curves so that
$$
\frac{\mathrm{d}}{\mathrm{d}t}(x,y)=(x,-y)\tag3
$$
which has the solution
$$
(x,y)=\left(x_0e^t,y_0e^{-t}\right)\tag4
$$
That is, $f$ is constant along the curves
$$
xy=x_0y_0\tag5
$$
Thus, $f(x,y)$ is a function of $xy$; so we can define $f(x,y)$ freely along $y=1$ and note that
$$
f(x,y)=f(xy,1)\tag6
$$
