Consider the implicit function given by the relation
$2xy-x^8+y^2=-2$
My problem is that $\dfrac{\mathrm dy}{\mathrm dx}$ turns out to be not definite (at least in my calculations) because it is $\dfrac{10}{0}$.
Could you please help me, finding the linear approximation in this case?
We have our function given by: $$2xy-x^8+y^2=-2$$ If we implicitly differentiate this, we obtain: $$2y+2x\frac{dy}{dx}-8x^7+2y\frac{dy}{dx}=0$$ We move all the terms without $\frac{dy}{dx}$ to the other side and factorise the terms with $\frac{dy}{dx}$: $$\frac{dy}{dx}(2x+2y)=8x^7-2y$$ $$\frac{dy}{dx}=\frac{4x^7-y}{x+y}$$ Indeed, you will realise that your tangent line will be at slope $\frac{dy}{dx} \to \pm \infty$ since the denominator tends to zero.
Therefore, your tangent (linear approximation) will be a vertical line.
Hence, look at your coordinates $(-1,1)$. Since the $x$-coordinate is given to be $x=-1$, that is the tangent line.
$$\text{Tangent line}: \boxed{x=-1}$$
If you find that $\tfrac{\mbox{d}y}{\mbox{d}x} \to \pm\infty$ in the given point, then the tangent line will be a vertical one.
If you feel uncomfortable with this, you could consider the equation as an implicit function $x(y)$ instead of $y(x)$. Then finding $\tfrac{\mbox{d}x}{\mbox{d}y}$ will give you a slope equal to $0$.
Alternatively, if you have covered functions of two variables, write the equation in the form $f(x,y)=0$ and then the tangent line at $(x_0,y_0)$ is also given by: $${\frac{\partial f}{\partial x}}\bigg\vert_{(x_0,y_0)}\left( x-x_0 \right)+{\frac{\partial f}{\partial y}}\bigg\vert_{(x_0,y_0)}\left( y-y_0 \right)=0$$