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Embed $Y$ in projective space and let $H$ be its hyperplane Cartier divisor. Since $f$ is surjective between integral surfaces, it is generically finite, with degree $r>0$. Its pullback of a Cartier divisor $L=f^{\ast }H$ is a nef divisor, and the projection formula for algebraic cycles gives $L^2=r{H}^2>0$. The same formula gives $L\cdot C=H\cdot f_{\ast }C=0$, because the curve is contracted to a point.

The Hodge index theorem for algebraic surfaces makes the intersection pairing on the Picard group of a surface have signature $(1,\rho -1)$ on $N^1(X{)}_{\mathbb{R}}$. The orthogonal complement of any positive-square vector, including $L$, is negative definite. The numerical class of $C$ is nonzero, since an ample divisor on $X$ intersects every nonzero effective divisor positively. Thus the contracted curve has negative self-intersection conclusion applies: