We have
$$\displaystyle{\left(1+x^2+y^2\right)^{\dfrac{1}{x^2+y^2+xy^2}}={\large{e}}^{\displaystyle{\left(\dfrac{\ln{ \left(1+x^2+y^2\right)}}{x^2+y^2+xy^2}\right)}}}.$$
We define $$f(x,y):= \frac{1}{x^2+y^2+xy^2} \cdot \ln{ \left(1+x^2+y^2\right)}.$$
We proceed with polar coordinates:
$$f\left(r \cos t, r \sin t\right)= \frac{1}{1+r \cos t \sin^2 t}\cdot \frac{\ln \left(1+r^2\right)}{r^2}.$$
Now we have
$$\displaystyle{\lim\limits_{r \to 0}\left(\frac{1}{1+r \cos t \sin^2 t}\right) \longrightarrow 1} \\ \qquad \\ \qquad \\ \text{and} \\ \qquad \\ \qquad \\ \lim\limits_{r \to 0}{\left(\frac{\ln \left(1+r^2 \right)}{r^2}\right)} \longrightarrow 1$$
Hence,
$$\lim\limits_{r \to 0}{f\left(r \cos t, r \sin t\right)} \longrightarrow 1 \\ $$
$$\displaystyle{\therefore \lim\limits_{(x, y) \to (0, 0)}{\left(1+x^2+y^2\right)^{\dfrac{1}{x^2+y^2+xy^2}}} \longrightarrow e}.$$