What is $f_{x}$ if $f(x, y)=x^{2} y^{2}+x^{y}$ ?
A. $4 x y+y x^{y+1}$
B. $2 x y^{2}+y x^{y-1}$
C. $2 x y^{2}+x^{y} \ln (y)$
D. $2 x y^{2}+1^{y}$
E. $2 x y^{2}+x^{y} \ln (x)$

Step 1 ：The question is asking for the partial derivative of the function \(f(x, y)=x^{2} y^{2}+x^{y}\) with respect to \(x\). The partial derivative of a function with respect to a variable is the derivative of the function with respect to that variable, treating all other variables as constants.

Step 2 ：To find the partial derivative of \(f(x, y)\) with respect to \(x\), we need to apply the power rule and the chain rule. The power rule states that the derivative of \(x^n\) is \(n*x^{n-1}\). The chain rule states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function.

Step 3 ：The derivative of \(x^{2} y^{2}\) with respect to \(x\) is \(2*x*y^{2}\), because \(y^{2}\) is treated as a constant.

Step 4 ：The derivative of \(x^{y}\) with respect to \(x\) is a bit more complicated. We need to use the chain rule. The outer function is \(u^{y}\), where \(u=x\). The derivative of \(u^{y}\) with respect to \(u\) is \(y*u^{y-1}\). The inner function is \(x\), and its derivative with respect to \(x\) is 1. So, the derivative of \(x^{y}\) with respect to \(x\) is \(y*x^{y-1}\).

Step 5 ：Adding these two derivatives together, we get \(f_{x}=2*x*y^{2}+y*x^{y-1}\).

Step 6 ：Final Answer: The partial derivative of the function \(f(x, y)=x^{2} y^{2}+x^{y}\) with respect to \(x\) is \(\boxed{2 x y^{2}+y x^{y-1}}\). So, the correct option is B. \(2 x y^{2}+y x^{y-1}\).