Answer
$$0$$
Work Step by Step
Given $$\lim _{(x, y) \rightarrow(0,0)} \tan \left(x^{2}+y^{2}\right) \tan ^{-1}\left(\frac{1}{x^{2}+y^{2}}\right) $$
Since
\begin{align*}
-\frac{\pi}{2} &\leq \tan ^{-1}\left(\frac{1}{x^{2}+y^{2}}\right) \leq \frac{\pi}{2}\\
-\frac{\pi}{2} \cdot \tan \left(x^{2}+y^{2}\right)& \leq \tan \left(x^{2}+y^{2}\right) \cdot\left(\frac{1}{x^{2}+y^{2}}\right) \leq \frac{\pi}{2} \tan \left(x^{2}+y^{2}\right)
\end{align*}
Then by using the Squeeze Theorem
$$\lim _{(x, y) \rightarrow(0,0)} -\frac{\pi}{2} \cdot \tan \left(x^{2}+y^{2}\right) \leq \lim _{(x, y) \rightarrow(0,0)} \tan \left(x^{2}+y^{2}\right) \cdot\left(\frac{1}{x^{2}+y^{2}}\right) \leq\lim _{(x, y) \rightarrow(0,0)} \frac{\pi}{2} \tan \left(x^{2}+y^{2}\right) $$
Then
$$0 \leq \lim _{(x, y) \rightarrow(0,0)} \tan \left(x^{2}+y^{2}\right) \cdot\left(\frac{1}{x^{2}+y^{2}}\right) \leq 0 $$
Hence $$ \lim _{(x, y) \rightarrow(0,0)} \tan \left(x^{2}+y^{2}\right) \cdot\left(\frac{1}{x^{2}+y^{2}}\right)=0 $$
