Is $f(x+iy)=x^2-y^2 + i\sqrt{|xy|}$ complex differentiable?

$$ f\;'(0) = \lim_{z\to0}\frac{f(z)-f(0)}{z-0} = \lim_{z\to0}\frac{f(z)}{z}.\tag{1} $$ In order that this limit exist, it is necessary that it be equal to the limit as $z$ approaches $0$ along any path that passes through $0$. Thus it must be the same as $$ \lim_{r\to0} \frac{f(re^{i\theta})}{re^{i\theta}}, $$ which is the limit in (1) as $z$ approaches $0$ along a certain line. The angle between that line and the $x$-axis is $\theta$. Since the function is expressed in terms of real and imaginary parts, write $x+iy=re^{i\theta}=r(\cos\theta+i\sin\theta)$. Then $$f(re^{i\theta})=f(x+iy)= x^2-y^2+i\sqrt{|xy|}=r^2(\cos^2\theta-\sin^2\theta) + i|r|\sqrt{|\cos\theta\sin\theta|}.$$ So $$ \frac{f(re^{i\theta})}{re^{i\theta}} = \frac{r(\cos^2\theta-\sin^2\theta) + i\operatorname{sign}(r)\sqrt{|\cos\theta\sin\theta|}}{e^{i\theta}} $$ where $\operatorname{sign}(r)= \pm1$ according as $r$ is positive or negative. So try taking the limit of that as $r\to0$.

Since $x^2−y^2$ is the real part of $z^2$, which is entire, the function cannot be analytic at (0,0) unless it agrees (in a 'hood of (0,0)) with $i2xy$, the imaginary part of $z^2$, which it does not, e.g., check along $y=x$