We consider the function $f:\mathbb{R}^2\rightarrow \mathbb{R}$ with $$f(x,y)=\frac{x^2-1}{y^2+1}$$
Describe and draw the level lines $N_c$ of $f$ for all $c\in \mathbb{R}$. Determine for each connected component of each non-empty level lines $N_c$ a parametrization $\gamma_c:I\rightarrow \mathbb{R}$.
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I have done the following :
We have that $$f(x,y)=c \Rightarrow \frac{x^2-1}{y^2+1}=c \Rightarrow x^2-1=(y^2+1)c \Rightarrow x^2-cy^2-(c+1)=0$$
If $c=0$ then we have $x^2=1 \Rightarrow x\pm 1$, so we get the lines $x=1$ and $x=-1$.
If $c>0$ we have then $c=m^2>0$. So $x^2-m^2y^2-(m^2+1)=0 \Rightarrow x^2-m^2y^2=(m^2+1)$. We have $B^2-4AC=0-4\cdot 1\cdot (-m^2)=m^2>0$. So it is the general equation of an hyperbola with axes parallel to the coordinate axes.
If $c<0$ we have then $c=-m^2<0$. So $x^2+m^2y^2-(-m^2+1)=0 \Rightarrow x^2+m^2y^2=(-m^2+1)$. The left side is always non-negative, so also the right side must be. So $1-m^2\geq0 \Rightarrow m^2\leq1 \Rightarrow -1\leq -m^2 \Rightarrow -1\leq c$. That means that is $c$ is negative then $c$ must be greater or equal to $-1$. If $-1<c<0$ then we have the general formula of an ellipse. If $c=-1$ then we get $x^2+y^2=0$ which is the point $(0,0)$.
Is that correct?
How do we determine for each connected component of each non-empty level lines $N_c$ a parametrization $\gamma_c:I\rightarrow \mathbb{R}$ ?
Your level lines are correct, confirmed by the following 3D representation of surface $z=f(x,y)$ (these contour lines are projected on the bottom plane).
About parametrization, use the classical:
$$x=a \cos(t), y=b \sin(t) \ \ \text{for ellipses and}\tag{1}$$
$$x=a \cosh(t), y=b \sinh(t) \ \ \text{ for hyperbolas}\tag{2}$$
For example, for the ellipse $(E_m)$ with equation :
$$x^2+m^2y^2=1-m^2 \ \iff \ \frac{x^2}{\left(1-m^2\right)}+\frac{y^2}{\left(\frac{1-m^2}{m^2}\right)}=1$$
by comparison with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, you get:
$$a=\sqrt{1-m^2}, \ b=\sqrt{\frac{1-m^2}{m^2}}$$
and use (1).
Same operation for hyperbolas.