Let $C$ be a curve in a smooth $3$-fold $X$ with an ordinary node $p\in X$. Blow-up $p$ let $E$ be the exceptional divisor, and $\widetilde{C}$ the strict transform of $C$. Furthermore let $L$ be the line in $E$ through the two points $E\cap \widetilde{C}$. 


Now, let us blow-up $\widetilde{C}$ (with exceptional divisor $E\_{C}$) and then flop the strict transform of $L$. Let $Z$ be the resulting variety. The strict transform of $E$ in $Z$ is isomorphic to $\mathbb{P}^1\times\mathbb{P}^1$, and can be blow-down to another variety $W$. We get then a divisorial contraction $f:W\rightarrow X$ contracting the strict transform of $E\_C$ to $C$.


Is $f$ the blow-up of $X$ along $C$ ? 


To me it is clear that it must be the blow-up over any point $q\in C\setminus \{p\}$ but I am quite confused on what happens over $p$. 



Is this divisorial contraction a blow-up?
The variety $Y$ you constructed is terminal. It is singular at the point where $\mathbb{P}^1\times\mathbb{P}^1$ is contracted. However this singularity is normal.


Now, you can write $C$ as $\{y^2-x^3-x^2=z=0\}\subset\mathbb{A}^3$. The blow-up $Z$ of $\mathbb{A}^3$ along $Z$ is defined by $\{s(y^2-x^3-x^2)-tz=0\}\subset\mathbb{A}^3\times\mathbb{P}^1$. Note that $Z$ has a unique singular point, namely $((0,0,0),[1:0])$ which is an ordinary double point. Therefore, $Z$ is terminal. 


Since there exists at most a unique terminal extraction of $C$ we get $Y\cong Z$.


