<s>[INST] Show that this relation is an implicit solution of the following differential equation Differential equation: 
$$\frac{dy}{dx}=\frac{xy}{x^2+y^2}$$
Relation:
$$2y^2 \ln{y} - x^2 = 0$$
From this, I end up getting: 
$$\frac{dy}{dx} = \frac{x}{2y\ln{y} + y} $$
The missing step would be to put in the form given at the top. How is this possible?
 [/INST] Multiply by $y/y$ first. Note from our relation $2y^2\log y-x^2=0$ that adding $x^2$ to both sides yields $2y^2\log y=x^2$. Substitute for $2y^2\log y$ and you are done.
$$\begin{align*}\frac{dy}{dx}&=\frac{x}{2y\log y+y}\\&=\frac{xy}{2y^2\log y+y^2}\\&=\frac{xy}{x^2+y^2}\end{align*}$$
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