Consider $y^8=x^{11}$.
So, $y^8-x^{11}=0$.
Define $f(x,y)=y^8-x^{11}$.
Then, $\nabla f = (f_x,f_y)=(-11x^{10},8y^7)$.
For $(f_x,f_y)=(0,0)$, we must have $(x,y)=(0,0)$; the singular point.
How do I go about resolving the singularity using blow ups? Are there any books with worked out examples?
The blow-up is locally given by $(u,v)\mapsto (u,uv)$ and $u=0$ is the exceptional divisor.
Hence, the pull-back of the curve $C$ given by $y^8-x^{11}$ is given by $(uv)^{8}+u^{11}=u^8(v^8+u^3)$. The pull-back is then equivalent to $8E+\tilde{C}$, where $C_1=\tilde{C}$ is the strict transform of the curve, given by $$u^3+v^8$$
Note that another chart for the blow-up is $(u,v)\mapsto (uv,v)$ but this one removes the point corresponding to the direction of $y=0$ so it is not good in your case, as it is exactly where your curve passes through.
The curve $C_1=\tilde{C}$ has a point of multiplicity $3$ with tangent $u=0$. You can then use $(u,v)\mapsto (uv,v)$ and obtain $(uv)^3+v^8=v^3(u^3+v^5)$ and the second strict transform $C_2=\tilde{C_1}=\tilde{\tilde{C}}$ has equation $$u^3+v^5.$$
Continuing this way you get $$u^3+v^2$$ $$u+v^2$$
Hence the sequence of multiplicities is $8$, $3$, $3$, $2$.