Answer
$\delta$ exists when $\delta =0.02$
Work Step by Step
We have $f(x,y)=\dfrac{x^3+y^4}{x^2+y^2}$ and $f(0,0)=0; \epsilon =0.02$
The polar-coordinates are defined as: $x= r \cos \theta , y = r \sin \theta$ and $r^2=x^2+y^2$
Consider $\sqrt {x^2+y^2} \lt \delta$
This implies that $\dfrac{|x+y|}{x^2+1} \lt 0.01$
Since,$|\dfrac{(r \cos \theta)^3+(r \sin \theta)^4}{x^2+y^2}| =| r \cos ^3 \theta +r^2 \sin^4 \theta| \lt 0.02$
When $\theta = \dfrac{\pi}{2} \implies |r^2| \lt 0.02$ [let us say $|r^2| =\delta$]
Thus, we can conclude that $\delta$ exists when $\delta =0.02$
