The graph above satisfy the equation $x^{4}+y^{3}=x^{2} y$.
The area enclosed by the 2 cute adorable little fine loops is equals to $\frac{a}{b}$ for coprime positive integers $a$ and $b$. What is the value of $a+b$ ?

Step 1 ：First, we rewrite the equation $x^{4}+y^{3}=x^{2} y$ as $x^{4}-x^{2} y+y^{3}=0$.

Step 2 ：Then, we can factor out $x^{2}$ to get $x^{2}(x^{2}-y)+y^{3}=0$.

Step 3 ：Setting each factor equal to zero gives us the equations $x^{2}=0$ and $x^{2}-y=0$ and $y^{3}=0$.

Step 4 ：Solving these equations gives us the points $(0,0)$, $(\sqrt{y},y)$, and $(0,0)$ respectively.

Step 5 ：These points are the intersections of the graph with the x-axis and y-axis, which are the vertices of the loops.

Step 6 ：We can find the area of each loop by integrating the function $y=x^{2}$ from $0$ to $\sqrt{y}$, and then doubling the result because there are two loops.

Step 7 ：This gives us the area $A=2\int_{0}^{\sqrt{y}}x^{2} dx$.

Step 8 ：Evaluating the integral gives us $A=2[\frac{1}{3}x^{3}]_{0}^{\sqrt{y}}=2\cdot\frac{1}{3}y^{\frac{3}{2}}$.

Step 9 ：Substituting $y=0$ into the equation gives us $A=2\cdot\frac{1}{3}\cdot0=0$.

Step 10 ：So, the area enclosed by the 2 loops is $0$, and the value of $a+b$ is $\boxed{0}$.