If you fix $\theta$ and just let $r\to 0^+$ then you are approaching $(0,0)$ only on straight lines. This can indeed be useful in order to show that a limit does not exist, i.e. providing two different values for $\theta$ which result in two different limits. If you want to cover every path that approaches $(0,0)$ and still use polar coordinates, then you need to consider $\theta$ as a function (e.g. $\theta=\theta(r)$ arbitrary), rather than a constant. In your example,
$$
\lim_{r\to 0^+}\frac{\big(r\cos\theta(r)\big)\cdot\big(r \sin\theta(r)\big)^2}{\big(r\cos\theta(r)\big)^2 + \big(r\sin\theta(r)\big)^2}
=
\lim_{r\to 0^+}\frac{r^3\cos\theta(r)\sin^2\theta(r)}{r^2}
=
\lim_{r\to 0^+}r\cos\theta(r)\sin^2\theta(r)
=
0
$$
Note that considering $\theta=\theta(r)$ rather than $r=r(s)$ and $\theta=\theta(s)$ with $r(s)\to0^+$ for $s\to 0^+$ is assuming you are somehow 'strictly approaching' $(0,0)$.
What you showed was that for each fixed $\theta\in [0,2\pi),$ the limit
$$\tag 1 \lim_{r\to 0^+} f(r\cos \theta, r\sin \theta) = 0.$$
A more informal way to say this: "$f$ has limit $0$ at the origin along every ray emanating from $(0,0).$"
THAT DOES NOT IMPLY $\,\lim_{(x,y)\to (0,0)} f(x,y) = 0.$ And it's crucial for you to understand this; otherwise limits in higher dimensions will be this weird vague mystery that forever makes you uneasy.
Let's look at $(1)$ more closely. In that process you are fixing $\theta$ and then letting $r\to 0^+.$ Yes, you get $0$ for the limit for every fixed $\theta,$ but that doesn't cut it. The definition of a limit as $(x,y)\to (0,0)$ doesn't involve looking only at rays, right? What about letting $(x,y)\to (0,0)$ when $(x,y)$ is on the parabola $y=x^2?$ Why would that be covered by the special case of rays? For the existence of a limit in this setting, the main idea is that when $(x,y)$ gets close to $(0,0),$ in any way whatsoever, $f(x,y)$ gets close to a limit $L.$
In fact in your problem we get
$$f(x,x^2) = \frac{x^2\cdot x^2}{x^4 + (x^2)^2} = \frac{1}{2}.$$
So $f$ is simply equal to $1/2$ at every point of the parabola $y=x^2$ where $x\ne 0.$ Since points on this parabola can be made as close to $(0,0)$ as we like, $f$ cannot have limit $0$ as $(x,y)\to (0,0).$ In fact $\lim_{(x,y)\to (0,0)} f(x,y)$ fails to exist (we get $1/2$ along the parabola and $0$ along each ray).
I think this specific problem is easier in rectangular coordinates, but if you want to think about it in polar coordinates, consider traveling to $(0,0)$ along the curve $\sin \theta = r$ as $\theta \to 0^+.$ Along this curve, which looks a lot like the parabola discussed above, we have
$$f(r\cos t, r\sin t) = \frac{r^2\cos^2 \theta\cdot r^2}{r^4\cos^4 +r^4} = \frac{\cos^2 \theta}{\cos^4\theta +1}.$$
As $\theta \to 0^+,$ the right side $\to 1/2,$ so again we see the limit of $f$ at the origin fails to exist.
Best Answer
You cannot take $\delta=\varepsilon-1$, since you are after a $\delta>0$.
And you don't need polar coordinates to show that that limit does not exist. Just use the fact that, if $f(x,y)=\frac{x^2+y}{\sqrt{x^2+y^2}}$, then