Show that the function $f:\mathbb{R}^2\rightarrow\mathbb{R}^2$ defined by $$f(x,y)=\frac{1}{10}(\sin^2(3x+y),\cos(4x+2y))$$ has a fixed point.
We have
$$D_1f_1(x,y)=\frac{6}{10}\sin(3x+y)\cos(3x+y)$$
$$D_2f_1(x,y)=\frac{2}{10}\sin(3x+y)\cos(3x+y)$$
$$D_1f_2(x,y)=-\frac{4}{10}\sin(4x+2y)$$
$$D_1f_2(x,y)=-\frac{2}{10}\sin(4x+2y)\mbox{.}$$
Apparently, the partials of $f$ are continuously differentiable on $\mathbb{R}^2$, and so is $f$.
Fix $(x_1,y_1),(x_2,y_2)\in\mathbb{R}^2$. Put
$$\phi(t)=(x_1+(x_2-x_1)t,y_1+(y_2-y_1)t)\qquad(0\leq t\leq 1)\mbox{.}$$
By the mean value theorem,
$$\lVert f(x_1,y_1)-f(x_2,y_2)\rVert=\lVert(f\phi)(0)-(f\phi)(1)\rVert\leq\lVert D(f\phi)(\xi)\rVert$$
for some $\xi$ lies between $0$ and $1$.
By the chain rule,
$$\begin{aligned}
\lVert D(f\phi)(\xi)\rVert&=\lVert Df(\phi(\xi))D\phi(\xi)\rVert\\&=\lVert(D_1f_1(x,y)(x_2-x_1)+D_2f_1(x,y)(y_2-y_1),D_1f_2(x,y)(x_2-x_1)+D_2f_2(x,y)(y_2-y_1)\rVert\\
&=\lVert(\frac{\sin(6x+2y)}{10}(3(x_1-x2)+(y_1-y2)),\frac{\sin(4x+2y)}{10}(-4(x2-x_1)-2(y_2-y_1))\rVert\\
&\leq\frac{\sqrt{2}}{10}\left(4\lvert x_2-x_1\rvert+2\lvert y_2-y_1\rvert\right)\\
&\leq \frac{4}{5}\lVert(x_1-x_2,y_1-y_2)\rVert\mbox{,}
\end{aligned}$$
where $x=x_1+(x_2-x_1)\xi$, $y=y_1(y_2-y_1)\xi$. Since $\mathbb{R}^2$ is complete, by the contraction mapping principle, $f$ has exactly one fixed point.
Does this look correct?