We have $\gamma:(-3,0)\to R^2$ map given by
$$\gamma(t)=(0,-t-2)$ $t\in(-3,-1]$$
$\gamma(t)$ any regular curve as given in the picture when $t\in (-1,-\frac{1}{\pi}]$
$\gamma(t)=\left(-t,\sin\frac{1}{t}\right)$ when $t\in\left(-\frac{1}{\pi},0\right)$
Can you kindly explain crucial steps how to get this graph?
I don't exactly which map you want to be plotted. I will do little generalization for you:
Suposse we have the map $\textbf{r}(t): t \in A \subseteq \mathbb{R} \rightarrow \mathbb{R^2} $ with equation: $$ \textbf{r}(t)=(x,y)=(f(t),g(t)) $$
The first step would be determining the range of $x$ and $y$ given the domain $A$ of the variable $t$.
The second step: Compute $\frac{dy}{dx}$. Let's assume $y=w(x)$. Then: $$ \frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=g'(t)\frac{dt}{dx} $$
The value $\frac{dt}{dx}$ can be computed using implicit differentation assuming $t=s(x)$: $$ 1=f'(t)\frac{dt}{dx}\rightarrow \frac{dt}{dx}=\frac{1}{f'(t)} $$
Finally: $$ \frac{dy}{dx}=\frac{g'(t)}{f'(t)} \quad \textbf{[1]} $$
The third step: Compute $\frac{d^2 y}{dx^2}$: $$ \frac{d^2 y}{dx^2}=\frac{d \left(\frac{dy}{dx}\right)}{dt}\frac{dt}{dx}= \frac{g''(t)f'(t)-g'(t)f''(t)}{\left(f'(t)\right)^3} \quad \textbf{[2]} $$
The fourth step: Find the roots in the domain $A$ of variable $t$ for the equations:
$$ \frac{dy}{dx}=\frac{g'(t)}{f'(t)}=0 \quad \textbf{[3]} $$ $$ \frac{d^2 y}{dx^2}=\frac{g''(t)f'(t)-g'(t)f''(t)}{\left(f'(t)\right)^3}=0 \quad \textbf{[4]} $$
The fifth step: Find the singularities of the function $\frac{dy}{dx}$ in the domain $A$ of variable $t$
The sixth step: With all the previous information, find where $\frac{dy}{dx}>0$, $\frac{dy}{dx}<0$, $\frac{d^2 y}{dx^2}>0$ and $\frac{d^2 y}{dx^2}<0$. Finally, proceed to plot.