The book Joyce: Riemannian Holonomy Groups and Calibrated Geometry contains on page 205 the Eguchi-Hanson space as an example:
Consider $\mathbb{C}^2$ with complex coordinates $(z_1,z_2)$, acted upon by the involution $-1:(z_1,z_2) \mapsto (-z_1,-z_2)$. Let $(X,\pi)$ be the blow-up of $\mathbb{C}^2/\{\pm 1\}$ at $0$. Then $X$ is a crepant resolution of $\mathbb{C}^2/\{\pm 1\}$. (...) Define $f:X \setminus \pi^{-1}(0) \rightarrow \mathbb{R}$ by $$ f= \sqrt{r^4+1}+2 \log r - \log \left( \sqrt{r^4+1} +1 \right), $$ where $r=\left( |z_1|^2 + |z_2 |^2 \right)^{1/2}$ is the radius function on $X$.
Define a $2$-form $\omega_1$ on $X \setminus \pi^{-1}(0)$ by $\omega_1=i \partial \overline{\partial} f$. Then $\omega_1$ extends smoothly and uniquely to $X$.
Question:
How can I see that $\omega_1$ extends to $X$?
Writing $r= \sqrt{z_1 \overline{z_1}+z_2 \overline{z_2}}$ I can compute $\omega_1$ explicitly. The expressions are long, so I used Mathematica. All partial derivatives $\frac{\partial}{\partial \overline{z_i}}\frac{\partial}{\partial z_k}f$ tend to $\infty$ as $r \rightarrow 0$. So it seems to me as if $f$ cannot be extended. What am I missing?
Eguchi-Honson should be singular at the origin of $\mathbb{C}^{2}$ and you have correctly found this fact by getting the limit $r \rightarrow 0$.
you have missed the point that you need to extend the metric over $\pi^{-1}(0)$ on $X$, not over $0 \in \mathbb{C}^{2}/\mathbb{Z}^{2}$!
To be more precise: the blow-up space $X$ is $\mathcal{O}_{\mathbb{CP}_{1}}(-2) \cong T^{*}\mathbb{CP}_{1}$ and $\pi^{-1}(0) = \text{Zero section of the bundle} \cong \mathbb{CP}_1 $. You need to check that "the metric is smoothly extendable over the zero section". Pullback $f$ with $\pi$ over $X/\pi^{-1}(0)$. By getting a local chart over a point of the zero section you will be able to check that $\partial \bar{\partial} \pi^{*}f$ can extend smoothly over the zero section.
The computation is subtle and using Mathematica would be useful.