$$f(x+iy) = \sin^2(x+y) + i\cos^2(x+y)$$
I have calculated
$$\frac{\partial u} {\partial x} = 2\sin(x+y)\cos(x+y)$$ $$\frac{\partial u} {\partial y} = 2\sin(x+y)\cos(x+y)$$ $$\frac{\partial v} {\partial x} = -2\cos(x+y)\sin(x+y)$$ $$\frac{\partial v} {\partial y} = -2\cos(x+y)\sin(x+y)$$
So we see that $\frac{\partial u} {\partial y}$ = -$\frac{\partial v} {\partial x}$ is already valid.
For the other CR equation it must hold $\frac{\partial u} {\partial x}$ = $\frac{\partial v} {\partial y}$ But here we have
$$2\sin(x+y)\cos(x+y) =^{!} - 2\sin(x+y)\cos(x+y)$$
Could I deduce from that they would only be valid if $x+y = 2\pi k$ or $x+y = \frac{\pi}{2} + \pi k$?
Would this be correct to say that the function would be nowhere holomorphic, because there is no neighborhood where it is complex differentiable only single points?
It does not change your final statement, but $2\sin(x+y)\cos(x+y)=- 2\sin(x+y)\cos(x+y)$ is not equivalent to "$x+y = 2\pi k$ or $x+y = \frac{\pi}{2} + \pi k$". It is equivalent to "$x+y = \pi k$ or $x+y = \frac{\pi}{2} + \pi k$", i.e. to $$x+y\in\frac\pi2\Bbb Z.$$
Note that you could find this condition a little more quickly: $$\begin{align}\overline\partial f(x+iy)=0&\iff\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}=0\\&\iff(1+i)(1-i)2\sin(x+y)\cos(x+y)=0\\&\iff\sin(2(x+y))=0\\&\iff2(x+y)\in\pi\Bbb Z. \end{align}$$ In your conclusion, I wouldn't say "only single points" (I don't know what it means) but "no interior points".