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.
As a complement of the other answers:

consider the function $f(x)=a^x+b^x$ with $a>0$ and $b>0$. Assume for simplicity that $a\neq 1$ and $b\neq 1$. There are a few cases to consider: 

1. If $a>1$ and $b>1$, then $a^x$ and $b^x$ are increasing, so $f(x)=a^x+b^x$ is increasing. As a consequence, $f$ is one-to-one so, for every constant $c$, the equation $f(x)=c$ has at most one solution. Precisely, the range of $f$ is $(0,\infty)$ so it has exactly one solution when $c>0$ and $0$ otherwise. 

This case applies to the equation $2^x+4^x=6$ which has only one solution, $x=1$.
2. Similarly, when $0<a<1$ and $0<b<1$, the function $f(x)=a^x+b^x$ is decreasing with range $(0,\infty)$, so every equation $f(x)=c$ has exactly one solution when $c>0$ and $0$ otherwise.
3. When $a>1$ and $0<b<1$, there are two or zero solutions to $f(x)=c$. The simple case is when $b=\frac{1}{a}$, since in that case, $f$ is even.

More generally, $f'(x)=\ln(a)a^x + \ln(b)b^x$. Since $\ln(a)>0$, $\ln(b)<0$, $a^x$ is increasing and $b^x$ is decreasing, then $f'$ is increasing. We have $\lim\_{x\to\pm\infty} f'(x)=\pm\infty$, so $f$ is decreasing on $(-\infty,\alpha]$, then increasing on $[\alpha,\infty)$, where $\alpha$ is the solution of $f'(x)=0$ (precisely, $\alpha=\ln\left(\frac{b}{a}\right)\ln\left(\frac{\ln(1/b)}{\ln(a)}\right)$).

So, if $c>f(\alpha)$, then the equation $f(x)=c$ has exactly two solutions.