Solution: We need to find compute \[ \lim_{(x,y) \to (0,0)} \left( \vert x \vert +\vert y \vert \right) \ln \left( x^4 + y^4 \right) . \] At first we recall the generalized mean inequality.. This inequality will be helpful to solve later problems also.

In our problem, we take $p=1$, $q=4$, and $w_1=w_2=\frac{1}{2}$, we get \begin{align*} & \left( \frac{1}{2}\cdot \vert x \vert + \frac{1}{2}\cdot \vert y \vert \right) ^{\frac{1}{1}} \leq \left( \frac{1}{2}\cdots |x|^4+ \frac{1}{2}\cdots \vert y \vert^4 \right)^{\frac{1}{4}} \\ \implies & \frac{\vert x \vert +\vert y \vert }{2} \leq \frac{1}{2^{\frac{1}{4}}} \left( x^4 + y^4 \right) ^{\frac{1}{4}} \\ \implies & \vert x \vert + \vert y \vert \leq \frac{2}{\sqrt[4]{2}} \sqrt[4]{x^4 + y^4} . \end{align*}

Therefore, we have \begin{align*} \left( \vert x \vert +\vert y \vert \right) \left\vert \ln \left( x^4 + y^4 \right) \right\vert & \leq \frac{2}{\sqrt[4]{2}} \sqrt[4]{x^4 + y^4} \left\vert \ln \left( x^4 + y^4 \right) \right\vert \\ & = \frac{2}{\sqrt[4]{2}} \sqrt[4]{x^4 + y^4} \left\vert 4\cdot \ln \left( x^4 + y^4 \right)^{\frac{1}{4}} \right\vert \\ & = \frac{8}{\sqrt[4]{2}} \sqrt[4]{x^4 + y^4} \left\vert \ln \sqrt[4]{x^4 + y^4} \right\vert. \end{align*} So we have \[ \lim_{(x,y) \to (0,0)} \frac{8}{\sqrt[4]{2}} \sqrt[4]{x^4 + y^4} \left\vert \ln \sqrt[4]{x^4 + y^4} \right\vert = 0, \] as \begin{align*} \lim_{x \to 0^+} x \ln x & = \lim_{x \to 0^+} \frac{\ln x}{\frac{1}{x}} \\ & = \lim_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{1}{x^2}} \quad \left( \text{Applying L'Hospital rule} \right) \\ & = \lim_{x \to 0^+} -x = 0. \end{align*}

NNow note that \[ 0 \leq \left( \vert x \vert +\vert y \vert \right) \ln \left( x^4 + y^4 \right) \leq \frac{8}{\sqrt[4]{2}} \sqrt[4]{x^4 + y^4} \left\vert \ln \sqrt[4]{x^4 + y^4} \right\vert, \] applying Sandwich theorem, we conclude that \[ \textcolor{blue}{\boxed{\lim_{(x,y) \to (0,0)} \left( \vert x \vert +\vert y \vert \right) \ln \left( x^4 + y^4 \right) = 0}} \]