Answer
$$ y'(x) =\frac{(4 x-1) \cdot \sqrt{x^{4}+y^{4}}-2 x^{3}}{\sqrt{x^{4}+y^{4}}+2 y^{3}}$$
Work Step by Step
Given $$2x^2-x-y =\sqrt{ x^4+y^4}$$
Differentiate with respect to $x$
\begin{align*}
\frac{d}{dx} \left(2x^2-x-y \right)&= \frac{d}{dx} \left(\sqrt{ x^4+y^4}\right)\\
4 x-1-y^{\prime}&=\frac{1}{2 \sqrt{x^{4}+y^{4}}} \cdot\left(4 x^{3}+4 y^{3} \cdot y^{\prime}\right)\\
\left[-1-\frac{2 y^{3}}{\sqrt{x^{4}+y^{4}}}\right] y^{\prime}&=\frac{2 x^{3}}{\sqrt{x^{4}+y^{4}}}+1-4 x\\
\frac{\sqrt{x^{4}+y^{4}}+2 y^{3}}{\sqrt{x^{4}+y^{4}}} y^{\prime}&=\frac{(4 x-1) \cdot \sqrt{x^{4}+y^{4}}-2 x^{3}}{\sqrt{x^{4}+y^{4}}}
\end{align*}
Then
$$ y'(x) =\frac{(4 x-1) \cdot \sqrt{x^{4}+y^{4}}-2 x^{3}}{\sqrt{x^{4}+y^{4}}+2 y^{3}}$$
