Resolution of the singularities of the curve $X^2Y^2 + Y^2Z^2 + X^2Z^2=0$ 
Problem: Let $C=\{[X,Y,Z] \in \mathbb{P^2}\mathbb{C} \mid X^2Y^2 + Y^2Z^2 + X^2Z^2=0\}$.
Resolve the singularities in $C$.

Tried so far: Since the polynomial is symmetric in the three variables, we can suppose wlog $Z=1$. Computing the derivatives we find out that the point $P=[0,0,1]$ is the only singularity on the curve of this kind, and thus there are 3 singularities, namely $P$, $[1,0,0]$ and $[0,1,0]$. We need to study $P$. The curve on the affine plane takes the form $x^2y^2 + y^2 + x^2=0$ with a singularity in the origin $(0,0)$. We attempt a blow up! Should we try $(x,y)=(u,uv)$ and $(x,y)=(uv,u)$ or just one of them? We try out the first one. We get $u^4v^2 + u^2v^2 + u^2=0$. Since $u$ is not $0$ (why??) we get $u^2v^2 + v^2 + 1=0$. Setting $u=0$ (we just said $u$ is not $0$!!! ... ???) we find two possible values for $v$ and thus the singularity should be a node. (are there any other possibilities?) We did it! There are 3 nodes on $C$. Is there a way to write down the equation for the curve with the singularities resolved?
Thanks!
 A: It looks like you did all the computations necessary, only that you're a bit confused as to how to interpret what you did.
1) In the chart $(u,v)$ as you used in your answer, the equation $u=0$ cuts out the exceptional divisor (the fiber of the blow up over the origin). The equation $u^4v^2+u^2v^2+u^2=0$ is what you get when you pull back $C$ to the blow up.
However, as you see from the equation, this equation also contains the exceptional divisor (which makes sense, as the original curve passed through the origin). This is called the total transform. 
However, we would like to remove these extraneous copies of the exceptional divisor. To do that, we divide the equation $u^4v^2+u^2v^2+u^2=0$ by as many copies of $u$ as possible. In this case, dividing by $u^2$ suffices, and we get the proper transform $u^2v^2+v^2+1=0$, which we can directly check is nonsingular.
2) Blowing up gives a recipe for desingularizing any plane curve. Any smooth curve can be mapped into $\mathbb{P}^2$ with only nodal singularities, but they cannot in general be embedded into $\mathbb{P}^2$, so it makes sense that blowing up doesn't give us an equation for the desingularized curve as a plane curve.
In our case, however, the arithmetic genus of a degree 4 curve is 3. Since there are three nodes, the genus of the desingularized curve is $3-3=0$. This means the desingularized curve is just a $\mathbb{P}^1$. 
