Answer
By Theorem 2, we conclude that $f\left( {x,y} \right) = {x^3}{y^8}$ is differentiable.
Work Step by Step
We have $f\left( {x,y} \right) = {x^3}{y^8}$. The partial derivatives are
${f_x}\left( {x,y} \right) = 3{x^2}{y^8}$, ${\ \ \ }$ ${f_y}\left( {x,y} \right) = 8{x^3}{y^7}$
Notice that ${f_x}\left( {x,y} \right)$ and ${f_y}\left( {x,y} \right)$ exist and are continuous. Therefore by Theorem 2, $f\left( {x,y} \right) = {x^3}{y^8}$ is differentiable.
