Suppose I have a homogeneous cubic polynomial $f(w,x,y,z)$ and I let $X$ be the set of points in $\mathbb{R}^4$ where $f(w,x,y,z)=0$ and $w^2+x^2+y^2+z^2=1$. Suppose that this is a smooth surface, and I give it the metric inherited from $\mathbb{R}^4$. Does anyone know a nice formula for the curvature of $X$ in terms of $w,x,y,z$ and the coefficients of $f$? Doubtless this could be worked out by going through the whole business of local coordinates and Christoffel symbols but I imagine that that would be quite lengthy and the final answer is probably not too complicated.
There is basic formula in Riemannian geometry that, used twice, gives fairly direct answer. Let $M$ be a Riemannian manifold, let $N$ be a codimension $1$ submanifold of $M$, and let $f$ be a function defined on $M$. Call $f$ also the restriction of $f$ to $N$. Then the Hessian of $f$ on $N$ and the restriction to $TN$ of the Hessian of $f$ on $M$ are related by $$ Hess_N(f) = Hess_M(f)+df(n) II~, $$ where $n$ is the oriented unit normal of $N$ in $M$ and $II$ is the second fundamental form of $N$ in $M$.
You can apply this first for $S^3$ considered as a hypersurface in $R^4$, and obtain that $$ Hess_{S^3}(f) = Hess_{R^4}(f)+(\partial_rf) g~, $$ where $g$ is induced metric on $S^3$. Then you can apply it to $X$ considered as a hypersurface of $S^3$ to obtain that $$ 0 = Hess_X(f) = Hess_{S^3}(f) + df(n) II~, $$ where $n$ is the unit normal to $X$ in $S^3$ and $II$ its second fundamental form.
Putting all this together yields a formula $$ II = (Hess_{R^4}(f) + (\partial_rf) g)/df(n)~. $$ The curvature of $X$ the follows from the Gauss formula, $K=1+det(II)$.

3$\begingroup$ Just one comment: since $X$ is a level set of $f$ in $\mathbb{S}^3$, $df(n)$ is nothing more than the norm of the projection of $df$ to $T^*S^3$. $\endgroup$ – Willie Wong Jul 9 '11 at 14:38