Let $f(x, y) = \begin{cases} \frac{2x^4y-5x^2y^2+y^5}{(x^2+y^2)^2}, \ (x, y) \neq (0, 0) \\ 0, \qquad \qquad \qquad (x, y) = (0,0 ) \end{cases}$. Find a $\delta > 0$ such that $|f(x, y)-f(0, 0)| < 0.1$ whenever $\sqrt{x^2+y^2} < \delta$.
Use polar coordinates: $$x=r\cos\theta, \qquad y=r\sin\theta$$ where it is given that $r < \delta$. Then the expression $f(r,\theta)$ becomes $$|f(x, y)-f(0, 0)| = |2r\cos^4 \theta\sin \theta +r\sin^5\theta -5\cos^2\theta\sin^2\theta| < 0.01$$ implies that $$-0.01 < 2r\cos^4 \theta\sin \theta +r\sin^5\theta -5\cos^2\theta\sin^2\theta < 0.01$$ $$ \implies 5\cos^2\theta\sin^2\theta-0.01 < 2r\cos^4 \theta\sin \theta +r\sin^5\theta < 5\cos^2\theta\sin^2\theta+0.01$$ $$ \implies \frac{5\cos^2\theta\sin^2\theta-0.01}{2\cos^4 \theta\sin \theta +\sin^5\theta} < r < \frac{5\cos^2\theta\sin^2\theta+0.01}{2\cos^4 \theta\sin \theta +\sin^5\theta}$$ where denominator is not $0$ and $\theta = \tan^{-1}\frac{y}{x}$.
If all the powers of $x$ and $y$ in the numerator add up to $5$, use polar coordinates: $$x=r\cos\theta\\y=r\sin\theta$$ Then you have an expression $f(r,\theta)=r g(\theta)$, where $g$ is a simple function. You can maximize it by taking both $\sin\theta$ and $\cos\theta$ to be $1$ (or $0$ or $-1$ if you need to subtract). Say if middle term is $-5x^3y^2$, I would use $g(\theta)<2\cdot1^4\cdot1+5\cdot1^3\cdot1^2+1^5=8$. The you have $f(r,\theta)<8r$. Now putting the condition $8r<0.1$ yields your desired $r$.