Rational singularity

Remarks

The condition does not depend on the chosen resolution. Since $X$ is normal and $f$ is proper and birational, one also has $f_* \mathcal{O}_Y = \mathcal{O}_X$, by Zariski's main theorem. Every nonsingular variety has rational singularities, with $f = \id$. Example: for the cone $X \colon xy = z^2$ in $\mathbb{A}^3$, the blow-up of the vertex is a resolution whose exceptional curve is $E \cong \mathbb{P}^1$ with normal bundle $\mathcal{O}(-2)$. Here $Y$ is the total space of the line bundle $\mathcal{O}(-2)$ on $E$, and $X$ is affine, so $R^1 f_* \mathcal{O}_Y$ is determined by $H^1(Y, \mathcal{O}_Y) = \bigoplus_{m \ge 0} H^1(\mathbb{P}^1, \mathcal{O}(2m)) = 0$. So the singularity is rational. Non-example: for the cone $X \colon x^3 + y^3 + z^3 = 0$ over a smooth plane cubic $E$, the blow-up of the vertex is the total space of $\mathcal{O}_E(-1)$, and the same computation gives $H^1(Y, \mathcal{O}_Y) = \bigoplus_{m \ge 0} H^1(E, \mathcal{O}_E(m))$. The term $m = 0$ is $H^1(E, \mathcal{O}_E) \cong k$, so $R^1 f_* \mathcal{O}_Y \ne 0$ and the singularity is not rational.