Counter-example: Cauchy Riemann equations does not imply differentiability
The function $f$ is analytic off the origin, hence satisfies the CR equations at all $z\ne0$. For $z=0$ look at $$f_x(0,0)=\lim_{x\to0}{e^{-1/x^4}\over x}=0\ ,$$ and similarly $$f_y(0,0)=\lim_{y\to0}{e^{-1/(iy)^4}\over y}=0\ .$$ It follows that $u_x(0,0)=v_x(0,0)=u_y(0,0)=v_y(0,0)=0$; so $f$ satisfies the CR equations also at $z=0$.
But $f$ is not even continuous at $z=0$: Consider the points $z(t):=(1+i)t$ for real $t$ near $0$. I leave the details to you.