Partial Derivative of $f(x,y) =\ln(x^{2} + y^{2}) + \sqrt{x^{2}\cdot y^{3}}$ $$f(x,y) = \ln(x^{2} + y^{2}) + \sqrt{x^{2}\cdot y^{3}}$$
What is the value of $f_{x}\left ( 0,1 \right )$ and $f_{y}\left ( 0,1 \right )$?
I tried but I found the denominator as zero.

$$f(x,y) = \ln(x^{2} + y^{2}) + \sqrt{x^{2}\cdot y^{3}}$$
Using chain rule of differentiation:
$$
\begin{align}
f_x(x, y) &= \dfrac{2x}{x^2 + y^2} + \dfrac{1}{2} 2xy^3 \left(x^2 \cdot y^3\right)^{\frac{-1}{2}} \\
&= \dfrac{2x}{x^2 + y^2} + \dfrac{xy^3}{\sqrt{x^2 \cdot y^3}} \\
&= \dfrac{2x}{x^2 + y^2} + y^{\frac{3}{2}}
\end{align}
$$
Similarly,
$$
\begin{align}
f_y(x, y) &= \dfrac{2y}{x^2 + y^2} + \dfrac{1}{2} 3x^2y^2 \left(x^2 \cdot y^3\right)^{\frac{-1}{2}} \\
&= \dfrac{2y}{x^2 + y^2} + 3xy^{\frac{1}{2}}
\end{align}
$$
