The exercise is to graph this function below.
$$y = \arctan\left(\frac{e^x-1}{\sqrt3}\right)-\arctan\left(\frac{e^x-4}{\sqrt3e^x}\right)$$
If the exercise were just one trigonometric function, then I would just put the expression inside arctangent in argtangent's domain and moreover put $y$ in the range of which the argtangent is defined. However there are two trigonometric functions in the equation above, and I do not know where I would start to able to sketch it.
I also want to add that I solved the equation algebraically by taking the tangent of both sides and using the subtaction of two angle for tangent
$$ \tan(x-y)=\frac{\tan\ x + \tan\ y}{1 + \tan\ x*\tan\ y}$$
The equation simplified to $\tan\ y = \sqrt3$, which gives us $y = \frac{\pi}{3}$. Does this $y$ value help me sketch the function in anyway?
Note that the arctan has an implied $+k\pi$ since $\tan(x+k\pi) =\tan(x)$, so you have to be careful. For example, if you naively evaluate $\arctan(1)+\arctan(2)+\arctan(3)$ you get $\arctan(1)+\arctan(2) =\arctan(\frac{1+2}{1-1\cdot 2}) =\arctan(-3) $ so $\arctan(1)+\arctan(2)+\arctan(3) =\arctan(-3)+\arctan(3) =0$, but the actual sum, using principal values, is $\pi/2$.
For this case, when $x > 0$,
$\begin{array}\\ y &= \arctan\left(\frac{e^x-1}{\sqrt3}\right)-\arctan\left(\frac{e^x-4}{\sqrt3e^x}\right)\\ &= \arctan\left(\frac{e^x-1}{\sqrt3}\right)-\arctan\left(\frac{1}{\sqrt3}-\frac{4}{\sqrt3e^{x}}\right)\\ &= t_1-t_2\\ \end{array} $
Since $e^x > 1$, $\frac{e^x-1}{\sqrt3} > 0$ so $0 < t_1 < \pi/2$ and $\frac{1}{\sqrt3}-\frac{4}{\sqrt3e^{x}} \gt \frac{1-4/3}{\sqrt3} =-\frac1{3\sqrt{3}} $ so $-\arctan(\frac1{3\sqrt{3}}) \le t_2 \lt \arctan(1/\sqrt{3}) =\pi/6 $.
Therefore $-\pi/6 \lt t_1-t_2 \lt \pi/2-\arctan(\frac1{3\sqrt{3}}) \approx \pi/2-0.190 $ and, since $\lim_{x \to \infty} t_1 = \pi/2$ and $\lim_{x \to \infty} t_2 = \pi/6$, we have $\lim_{x \to \infty} t_1-t_2 = \pi/3$.
For $x < 0$ you can do a similar analysis using $0 < e^x < 1$.