It is mentioned in this question https://mathoverflow.net/questions/148826/do-there-exist-double-points-on-an-algebraic-surface-in-mathbbp-mathbbc that $X=\{x^2+y^3+z^6=0\}\subset \mathbb C^3$ defines an elliptic singularity at the origin. This means that there is a desingularization $$f:\tilde{X}\to X$$ such that the curve over $0$ has the arithmetic genus $p_a(C_0)=1$.
In page 109, exercise 18 of Miles Reid's notes, weighted blowup is suggested to resolve the singularity.
Edited: As @Sasha pointed out, the singularity can also be resolved just by a sequence of ordinary blowups, so let's try it: Blowup the origin gives the system of equations
$$x^2+y^3+z^6=0,xv=yu,xw=zu,yw=zv$$
in $\mathbb C^3\times \mathbb P^2_{[u,v,w]}$. By setting $v=1$ and substitute, we get $y^2(u^2+y+y^4w^6)=0.$ Therefore the proper transform is defined by $u^2+y+y^4w^6=0$ in the affine chart and it is smooth, moreover the exceptional divisor is the rational curve $\{u=0\}\subset \mathbb P^2_{[u,v,w]}$ with multiplicity $2$.
However, I didn't see an elliptic curve appear in the blowup. Is there anything wrong?
Thanks!