Answer
\begin{align*}
(a)&\frac{\partial f}{\partial x} = 15x^2 +6xy^3+9y+14\\
(b)&\frac{\partial f}{\partial y} =9x^2y^2+9x\\
(c)&\frac{\partial f}{\partial x}\bigg|_{y=2} = 15x^2 +48x+86\\
\end{align*}
Work Step by Step
Given $$f(x, y)=5 x^{3}+3 x^{2} y^{3}+9 x y+14 x+8$$
Since
\begin{align*}
(a)&\frac{\partial f}{\partial x} = 15x^2 +6xy^3+9y+14\\
(b)&\frac{\partial f}{\partial y} =9x^2y^2+9x\\
(c)&\frac{\partial f}{\partial x}\bigg|_{y=2} = 15x^2 +48x+86\\
\end{align*}
