# Prove that if $f(x,y)$ is homogeneous of order $0$, then $f(x,y)=g\left(\frac{y}{x}\right)$, for some function $g$.

I often see in the case of homogeneous differential equations that if $$f(x,y)$$ is a homogeneous function of order $$0$$ and, $$\frac{\mathrm{d}y}{\mathrm{d}x}=f(x,y),$$ then the substitution $$y=ux$$ for some function $$u$$ TBD, turns it into a separable differential equation, this is easy to prove. However, then they proceed to say $$f(x,y)$$ must be of the form $$g\left(\frac{y}{x}\right)$$, for some function $$g$$ without any proof.

Sure, if $$f(x,y)=g\left(\frac{y}{x}\right)$$ then of course it is homogeneous of order $$0$$, but how do we know that this is the only case? I've tried contrapositive proofs, contradictions but none of them seemed to work.

Define $$g(z) = f(1, z)$$.

Then, ignoring points for which $$x=0$$,

\begin{align} f(x, y) & = f(x \cdot 1, x \cdot \frac{y}{x}) \\ & = f(1, \frac{y}{x}) \text{ using order 0 homogeniety}\\ & = g(\frac{y}{x}) \end{align}

We have that

$$f(\lambda x, \lambda y) = f(x,y)$$

Take the derivative w.r.t. $$\lambda$$ on both sides

$$xf_x(\lambda x, \lambda y) + yf_y(\lambda x, \lambda y) = 0$$

since this is true for all values of $$\lambda$$, set $$\lambda = 1$$

$$xf_x+yf_y = 0$$

In polar coordinates, using $$x\partial_x + y\partial_y = r\partial_r$$ we get the differential equation

$$f_r = 0 \implies f(r,\theta) = g(\theta)$$

and $$\theta$$ is a function of $$\frac{y}{x}$$