Let $X\subset{\mathbb{C}}^{2}$ be the irreducible cubic with equation $xy=x^{3}+1$. Then

$$\overline{X}=\{[t:x:y]\in{\mathbb{P}}^{2}:\,xyt=x^{3}+t^{3}\},\;\overline{X}=X\cup\{a\},\;a=[0:0:1].$$

The germ $(\overline{X},a)$ has two irreducible components $X_{1},\,X_{2}$, both are smooth at $a$, $X_{1}$ being tangent to the line $\{x=0\}$, and $X_{2}$ to the line $\{t=0\}$.

Note that in fact $X\subset{\mathbb{C}}^{\star}\times{\mathbb{C}}$ is the graph of the rational function $y=x^{2}+x^{-1}$, $x\in{\mathbb{C}}^{\star}$. If $(x,y)\in X$ and $x\to 0$ then $(x,y)\to a$ along $X_{1}$, while as $x\to\infty$ then $(x,y)\to a$ along $X_{2}$. The function

$$u(x,y)=\max\{-\log|x|,2\log|x|+1\}$$

is psh in ${\mathbb{C}}^{\star}\times{\mathbb{C}}$. It is easy to check that $\eta:=u\,|_{{}_{X}}\in{\mathcal{L}}(X)$ and

$$\limsup_{X_{1}\ni[1:\zeta]\to a}(\eta(\zeta)-\rho(1,\zeta))=0\;,\;\;\limsup_{X_{2}\ni[1:\zeta]\to a}(\eta(\zeta)-\rho(1,\zeta))=1.$$

Hence $\eta$ does not admit an extension in ${\mathcal{L}}({\mathbb{C}}^{2})$.