Find the continuous function $f$ with domain $\mathbb C$ such that it's differentiable on $x^2-y^2=1$ but is not differentiable at other points of $\mathbb C$.
I assume $f=u(x,y)+iv(x,y)$, all points on $x^2-y^2=1$ are in the form $(x,\sqrt{x^{2}-1})$ or $(x,-\sqrt{x^{2}-1})$, So if $f$ is going ot be differentiable at $(x,\sqrt{x^{2}-1})$ then $$u_x(x,\sqrt{x^{2}-1})=v_y(x,\sqrt{x^{2}-1})$$ $$u_y(x,\sqrt{x^{2}-1})=-v_x(x,\sqrt{x^{2}-1})$$
And for points in the form $(x,-\sqrt{x^{2}-1})$ $$u_x(x-,\sqrt{x^{2}-1})=v_y(x,-\sqrt{x^{2}-1})$$ $$u_y(x,-\sqrt{x^{2}-1})=-v_x(x,-\sqrt{x^{2}-1})$$
How does that help to find $f$?
We want a function, $f(x,y)= u(x,y)+v(x,y)i$ such that $u_x=v_y$ and $u_y=-v_x$ on the unit circle and fails those equations off of the unit circle. Well, if I want $u_x=v_y$ when every $x^2+y^2=1$, then lets try to manipulate the $x^2+y^2=1$ to fit that equation, maybe $x^2=1-y^2$? So we see to set $u_x = x^2$ and $v_y = 1-y^2$. We then arrive at
$$f(x,y) = \frac{1}{3}x^3 + \left(y-\frac{1}{3}y^3\right) i$$
Double check that this satisfies the Cauchy Riemann equations for every point on the unit circle (i.e., $x^2+y^2=1$), but fails for every point that is not on the unit circle (i.e., $x^2+y^2\not=1$). This follows pretty easily since the second equation $u_y=-v_x$ is trivially $0=0$, so you only have to check the first equation.
Note: The word "the" in the phrase "the continuous function" is incorrect. There are clearly many continuous functions that work.