Converting a differential equation 
Consider an ODE $\frac{dy}{dx}=h(x,y)$ such that $h(rx,ry)=h(x,y)$, which implies $h(x,y)=k\left(\frac{x}{y}\right)$.

(Why?)

Show that this ODE can be changed to a separable ODE for $u=u(x)$, if $u=\frac{y}{x}$.

This question is from Michael E. Taylor, Partial Differential Equations, Part 1. (Exercise 1, of section 1.7).
 A: Use $r=\frac{1}{x}$ then $$y'=h(x,y)=h\left(\frac{1}{x}x,\frac{1}{x}y\right)=h\left(1, \frac{y}{x}\right)=k\left(\frac{y}{x}\right)$$
now use $u=\frac{y}{x}\Rightarrow y'=u'x+u$ hence $u'x+u=k(u)$ which is a separable ODE.
A: The condition $$h(rx, ry) = h(x, y)$$ (which probably is meant to hold for all $r \neq 0$, lest $h$ necessarily be constant) says precisely that $h$ is constant on the lines parameterized by $r \mapsto (rx, ry)$, that is the lines through the origin. On the other hand, any such line is determined by its slope, namely, the ratio $\frac{y}{x}$ for any point $(x, y)$ on the line other than the origin. Thus, $h(x, y)$ is some function of $\frac{y}{x}$ (or more precisely, the restriction of $h$ to $\{x \neq 0\}$ is such a function).
We can think of the spirit of the problem geometrically: Ignoring the origin (which we ought to do anyway, as any function $h$ continuous at the origin and satisfying the above identity must be cosntant), the condition says that $h$ descends to a function $K$ on $\Bbb R^2 - \{ 0 \} / \sim$, where $\sim$ denotes the equivalence relation where $(x, y) \sim (x', y')$ if the two points lie on the same line through the origin. But this is just the projective line $\Bbb P^1$, and $k\left(\frac{x}{y}\right)$ is just the coordinate representation of $K$ in the standard affine chart on $\Bbb P^1$ with chart map $[x, y] \mapsto \frac{x}{y}$.
A: First let's answer you why point.
Define $k(x)=h(x^2,x)$. For $(x,y) \in \mathbb R^2$ with $y \neq 0$, you have $$h(x,y)=h\left(\frac{x}{y^2}x,\frac{x}{y^2}y\right)=h\left(\left(\frac{x}{y}\right)^2,\frac{x}{y}\right)=k\left(\frac{x}{y}\right)$$
Now your ODE is $$\frac{dy}{dx}=k\left(\frac{x}{y}\right)$$ and $$\frac{dy}{dx}=u+\frac{du}{dx}=k(u)$$
