I have to prove that there are two numbers $x_1,x_2 \in (-1,1)$ that are the unique solution of the system $\{3x=\sin(x+y+13), 3y=\cos(x+y)\}$.
Considering $x_1,x_2 \in (\frac{-1}{3},\frac{1}{3})$ we can put the system in two 1-variable equations:
$x=\frac{1}{3}\sin(\frac{\cos(13)}{3}\sqrt{1-9x^2}+(1+\sin(13))x+13)=:f(x)$
$y=\frac{1}{3}\cos(\frac{\cos(13)}{3}\sqrt{1-9y^2}+(1+\sin(13))y)=:g(y)$
I don't know how to continue from this point because I'd have to prove that exist $L_1, L_2 \in [0,1)$ such that $|f'(x)|\leq L_1$ and $|g'(y)|\leq L_2$ for all $x,y\in (\frac{-1}{3},\frac{1}{3})$, then conclude that $f$ and $g$ have an unique fixed point, respectively, and those points are the solution of the main system.
My problem is working with the derivative functions to obtain the $L_1, L_2$, I think it's too hard bounding them.
Let $f(x,y)=\left(\frac{1}{3}\sin(x+y+13),\frac{1}{3}\cos(x+y)\right)$ for $(x,y)\in [-1,1]\times[-1,1]=A$. Then $f$ maps $A$ to $A$, and for $x_1,x_2,y_1,y_2\in[-1,1]$, \begin{align*}\|f(x_1,y_1)-f(x_2,y_2)\|^2&=\frac{1}{9}\left(\sin(x_1+y_1+13)-\sin(x_2+y_2+13)\right)^2\\ &\qquad+\frac{1}{9}\left(\cos(x_1+y_1)-\cos(x_2+y_2)\right)^2\\ &\leq \frac{1}{9}|x_1+y_1-x_2-y_2|^2+\frac{1}{9}|x_1+y_1-x_2-y_2|^2\\ &\leq\frac{4}{9}\left(|x_1-x_2|^2+|y_1-y_2|^2\right),\end{align*} since $(a+b)^2\leq 2a^2+2b^2$ for all $a,b$. Therefore, for any $P,Q\in A$, $$\|f(P)-f(Q)\|\leq\frac{2}{3}\|P-Q\|,$$ hence $f$ is a contraction in $A$. Therefore $f$ has a unique fixed point in $A$.