Where is $f(x + y i) = x^2-y^2$ complex differentiable? The function $f$ is defined on the complex numbers by $f(x + y i) = x^2 - y^2$, where $x$ and $y$ are the real and imaginary parts of the parameter.
Our class's notes say that $f$ is complex differentiable along the line $y=-x$, but I keep getting by Cauchy Riemann equation that the origin is the only possible point. Am I missing something here? $f_x = 2x$ and $f_y = -2y$ so by $f_x= -i*f_y$ only when $x = 0 = y$...

My guess is that what your notes say is that if $f(x+yi)=x^2-y^2i$, then the points of $\mathbb C$ at which $f$ is diferentiable are those of the form $x-xi$ (and only those). This follows indeed from the Cauchy-Riemann equations, since, in this case, $u(x,y)=x^2$ and $v(x,y)=-y^2$.
