Let $$x^2+xy+y^2=2$$ Find $\displaystyle\frac{dy}{dx}$
Applying $\frac{d()}{dx}$ on both sides gives $$\frac{dy}{dx}(2y+x)=-(2x+y)$$
For $x\neq -2y$
$$\frac{dy}{dx}=-\frac{(2x+y)}{(2y+x)}$$
Substituting $x=-2y$ in original equation gives two solutions. Let them be $(\alpha,\beta) \text{&} (\gamma,\delta)$
Now clearly we can't use above definition to calculate derivative at $x=\alpha,\gamma$
I thought that following limit can help
$$\lim_{x\to \alpha}\frac{dy}{dx}=f'(\alpha)$$
since for points near to $(\alpha,\beta)$, we can use the derivative definition we found above.
Is this correct ?
Your approach relies on the fact that the derivative $\frac{dy}{dx}$ is a continuous function of $x$. But you can avoid (explicitly) taking limits by considering the reciprocal-slope $\frac{dx}{dy}$. When you implicitly differentiate with respect to $y$, you will get$^\dagger$ the reciprocal expression: $$ \frac{dx}{dy} = - \frac{2y+x}{2x+y} $$
This is clearly equal to $0$ when $x=-2y$. In the standard coordinates, with $x$ horizontal and $y$ vertical, this means that the tangent lines at those points are vertical, i.e. the heuristic calculation $$ \frac{dy}{dx} \mathrel{``="} \frac{1}{0} \mathrel{``="} \infty $$ is correct.
Here's a sketch, including the two points on the ellipse where $x=-2y$, namely $$ (x, y) = \biggl( \pm\frac{2\sqrt6}{3}, \mp\frac{\sqrt6}{3} \biggr). $$
$^\dagger$ You don't have to literally go through the steps of differentiating with respect to $y$. You can just write down the reciprocal expression since by the chain rule, $$ \frac{dy}{dx} \frac{dx}{dy} = 1. $$