Which isolated surface singularity comes from a -5 curve? Define the surface $X$ to be the total space of $\mathcal{O}_{\mathbb{P}^1}(-5)$.
By contracting the exceptional curve in $X$, we get a surface with an isolated singularity. I am looking for the equation (or the set of equations) that describes this singularity (as a surface in some $\mathbb{C}^n$, possibly just $\mathbb{C}^3$).
For example in the case of $X$ being the total space of $\mathcal{O}_{\mathbb{P}^1}(-2)$,
the resulting singularity is the $A_1$ singularity given by
$$x^2+yz=0$$.
 A: Let $C\subset\mathbb{P}^n$ be a rational normal curve of degree $n$, and $X$ the cone over $C$ with vertex $p\in\mathbb{P}^n$. Blowing up $p$ in $X$ you get the Hirzebruch surface $\mathbb{F}_n = \mathbb{P}(\mathcal{O}\oplus\mathcal{O}(-n))$. The exceptional divisor $E\subset\mathbb{F}_n$ is a genus zero smooth curve with $E^2 = -n$.
Therefore blowing down $E$ you get the vertex of a cone over a rational normal curve of degree $n$, that is a Du Val singularity of type $A_n$. This singularity is given by the equation
$$x^2+y^2+z^{n+1}= 0.$$
In your specific case $n = 5$ you have the vertex of a cone over a rational normal curve of degree $5$ in $\mathbb{P}^{5}$ given by $x^2+y^2+z^6 = 0$.
As Jérémy Blanc wrote these singualrities can be viewed as the rational quotient singularities of type $\frac{1}{n}(1,1)$ of the weighted projective plane $\mathbb{P}(1,1,n)$ at $[0:0:1]$.
A: The weighted projective plane $\mathbb{P}(1,1,n)$ can be viewed as $\mathbb{P}(1,1,n)=\mathbb{C}^3\setminus \{0\}/(x,y,z)\sim (\lambda x,\lambda y,\lambda^n z)$. For $n=1$ we obtain the standard projective plane. For $n>1$, the point $(0,0,1)$ is the unique singular point, and the blow-up of this point is a Hirzebruch surface $\mathbb{F}_n$, with exceptional divisor $E\simeq \mathbb{P}^1$ of self-intersection $-n$. Hence, what you are looking for is just the quotient singularity of $\mathbb{P}(1,1,n)$.
If you want a local embedding into some affine space, take the local embedding $(x,y,z)\mapsto (x^n/z,x^{n-1}y/z,\dots,xy^{n-1}/z,y^n/z)$. The equations are probably easy to obtain from this explicit description. For $n=2$ you obtain what you already described.
