So, I was adviced to ask a new question on my problem (as the first one wasn't very precise), that is to solve the system of equations: $$\begin{cases} x\cdot y=6 \\ x^y+y^x=17 \end{cases}$$ where:
- solutions are considered up to symmetry,
- $x,y\in \{z\in\Bbb R\mid z\geq 0\}$
I'm interested in methods different from the Newton's method or the listing that if $xy=6$, then $x=2, y=3$ or $x=1,y=6$ (and so on) and then manually checking whether such pair satisfies $x^y+y^x=17$.
From the first equation we have $$y=\frac6x.$$ Plugging this into the second equation yields $$x^{\tfrac6x}+\left(\frac6x\right)^x=17.$$ Call this expression on the left hand side $f(x)$. We want to solve the equation $f(x)=17$. It is obvious that $x=2$ and $x=3$ solve this equation. It remains to show that there are no other solutions.
To see this, first rewrite $f$ as $$f(x)=e^{\frac6x\log x}+e^{x\log\frac6x}.$$ Computing the derivatives of $$g(x)=\frac6x\log x\qquad\text{and}\qquad h(x)=x\log\frac6x$$ immediately shows that $g$ is strictly increasing on $(0,e]$ and strictly decreasing on $[e,\infty)$. Similarly, $h$ is strictly increasing on $(0,\frac6e]$ and strictly decreasing on $[\frac6e,\infty)$. Since the exponential function is increasing, we may conclude that $f$ is increasing on $(0,\frac6e]$ and decreasing on $[e,\infty)$.
Therefore $f(x)<17$ for $x\in(0,2)\cup(3,\infty)$ and $f(x)>17$ for $x\in(2,\frac6e]\cup[e,3)$.
We now only have to verify that $f(x)>17$ for $x\in[\frac6e,e]$. Note that $g$ is increasing on $[\frac6e,e]$, so $g(x)\geq g(\frac6e)=e\log(\frac6e)$ for $x\in[\frac6e,e]$. Similarly, $h$ is decreasing there, so $h(x)\geq h(e)=e\log(\frac6e)$ for $x\in[\frac6e,e]$. Therefore, for $x\in[\frac6e,e]$ we have $$h(x)\geq e^{e\log(\frac6e)}+e^{e\log(\frac6e)}=2\left(\frac6e\right)^e>17.$$
Therefore the only solution is $\{x,y\}=\{2,3\}$.