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: 
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.
The term "solve" generally means a systematic method of finding solutions, and showing that they are exhaustive (that is, no further solutions exist). This method fails on both counts. 

First, while for the particular example you give, it's not too difficult to see that 1 is a solution, I don't see any way to generalize this to a general method of solving equations of this type. It appears to just be guess and check, and while guess and check can be considered to be a method of solving in the broadest sense, it's not what people usually mean by the term.

Second, this doesn't show that 1 is the only solution. There are cases where you have $x^a+y^a = x^b+y^b$ but $a \neq b$. One situation where that can happen is if $y = x^{-1}$. Then $x^{-a}+y^{-a} = (x^{-1})^a +(y^{-1})^a = y^a+x^a = x^a+y^a$. In fact, your original statement that $(x^a = x^b) \rightarrow (a = b)$ holds only for $x$ positive real and $a, b$ real.