= Solution
For an irreducible <algebraic variety> $X$ and $p\in X$, the local ring $\mathcal O_{X,p}$ consists of germs of <rational function>[rational functions] regular on a neighbourhood of $p$. If $X\subseteq\mathbb A^n$ has ideal $I(X)$, its <Zariski tangent space> is
$$
T_pX=\{v\in k^n:df_p(v)=0\text{ for every }f\in I(X)\}.
$$
The given variety is the <blowup of the affine plane at the origin>. On the chart $W\ne0$, put $W=1$; then $Y=XZ$, so $(X,Z)$ are free affine coordinates. On the chart $Z\ne0$, put $Z=1$; then $X=WY$, so $(W,Y)$ are free affine coordinates. These two smooth affine-plane charts cover $V$, hence every point of $V$ is smooth.
If $(X,Y)\ne(0,0)$, the equation forces
$$
[W:Z]=[X:Y].
$$
Consequently $\pi$ restricts to an isomorphism away from the origin, and is therefore <birational variety>[birational]. Above the origin, however,
$$
\pi^{-1}(0,0)=\{(0,0)\}\times\mathbb P^1,
$$
so $\pi$ is not injective and cannot be an <isomorphism of algebraic varieties>. Birationality gives
$$
\boxed{k(V)=k(\mathbb A^2)=k(X,Y)}.
$$
For any morphism $\varphi:V\to V'$ with $V'$ affine, its restriction to the exceptional curve $E\cong\mathbb P^1$ is constant: every regular function on a projective line is constant, and the affine coordinate functions of $V'$ therefore have constant pullbacks. Since $E$ has more than one point, $\varphi$ is not injective. If $V$ itself were affine, its identity morphism would contradict this conclusion, so $V$ is not affine.
Over $\mathbb C$, $V$ is not compact in the Euclidean topology: the sequence $(n,0,[1:0])$ has no convergent subsequence. Every complex <projective variety> is Euclidean compact. Since <compact space>[compactness] is preserved by homeomorphisms, $V$ cannot be homeomorphic to a projective variety.
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