Question: I have to find $u(x,y)$ and $v(x,y)$ such that $f=u+iv$ is analytic. The textbook says that: $$2xyu+(y^2-x^2)v+2xy(x^2+y^2)^2=0$$

I've already tried to divide all the relation by $2xy$ (supposing $x$, $y$ non-zero) and then calculating the second order partial derivatives (to prove that $u$ or $v$ are harmonic). The computations got pretty complicated, so I was wondering, is there an easier way to find $u$ and $v$?

Answer: 
No.

It's better:
$$(2^x)^2+2^x-6=0$$ or
$$(2^x-2)(2^x+3)=0$$ or $$2^x=2,$$ which gives $$x=1.$$
Also, you can say that $2^{2x}+2^x$ increases, which by your work gives $x=1$ again.
Well, this is one way to obtain a solution, especially if you don't know any other way to proceed -- we usually say the solution has been obtained *by observation.*

However, the drawback with this method is that it may not work with a sufficiently complicated equation. Also, it doesn't guarantee that one has found all the solutions -- although one may ascertain this by other means.

In this case you can solve in a more sure way, as one of the answers has demonstrated, by noticing that this is quadratic in $2^x.$

But once again, observation is one way to proceed when nothing else works.