Let $D$ be the projective curve defined by $y^2z = x^3.$ Consider the map $f: \mathbb{P}_1 \to D$ defined by$$f[s, t] = [s^2t, s^3, t^3].$$Is it necessarily a homeomorphism? Any help would be greatly appreciated.
Progress so far: I think that $f^{-1}[x, y, z] = [y, x] = [\sqrt{x}, \sqrt{z}]$ is valid step after choosing a branch?
I assume you are working over $\Bbb{R}$, in which case there are no choices of branches involved. We will show that the map $$ g[x,y,z] = [y^{1/3},z^{1/3}] $$ is an inverse for $f$. Indeed, we have $$ (g \circ f)[s,t] = g[s^2t,s^3,t^3] = [(s^3)^{1/3}, (t^3)^{1/3}] = [s,t] $$ and if $[x,y,z] \in D$ then $$ (f \circ g)[x,y,z] = f[y^{1/3},z^{1/3}] = [y^{2/3}z^{1/3}, y, z] = [x,y,z] $$ because $x^3 = y^2z$ implies $x = y^{2/3} z^{1/3}$.
Finally, observe that both $f$ and $g$ are continuous, because their components are. Thus $f$ is a homeomorphism, as required.
As a side-note, observe that $D$ is a cusped cubic. This means that it cannot be diffeomorphic to $\Bbb{P}^1$.