Show that the projective closure of V(x) and V($x-y^4-z^4$) in $P^3$ is not isomorphic This is exercise 3.4.4 from "An Invitation to Algebraic Geometry": Show that the affine varieties V(x) and V($x-y^4-z^4$) is isomorphic in $A^3$ but their projective closures are not in $P^3$.

What I did:

They're isomorphic in $A^3$ since they're both isomorphic to $A^2$, and an isomorphism between them is $[0;y;z] \mapsto [y^4+z^4;y;z]$. Then I tried to show that the projective varieties V(x) and V($w^3x-y^4-z^4$) is not isomorphic in $P^3$. The first one is now a $P^2$, the second one is also topologically a $P^2$ so I'm trying to find somewhere the second one is not smooth but failed. Could anybody tell me a point where the second one is not smooth?

For a projective hypersurface $V(f) \subset \mathbb P^n$ defined by a homogenous polynomial $$f(x_0,...,x_n)$$ there is a simple criterion to characterize a singular point of $V(f)$: The point $x = (x_0:\ ...\ :x_n) \in \mathbb P^n$ is a singular point $x \in V(f)$ iff all partial derivatives vanish, i.e. 
$$\frac {\partial f}{\partial x_i}(x) = 0, i = 0,...,n.$$
Note. Euler's Lemma (see Hartshorne Ex. I,5.8) implies that $x \in V(f)$.
