Find points on the curve $\mathrm{y}={\mathrm{x}}^3-3{\mathrm{x}}^2-4\mathrm{x}$ at which the tangent lines are parallel to the line $4\mathrm{x}+\mathrm{y}-3=0$

First, we will find the slope of tangent,
$\begin{array}{l}\mathrm{y}={\mathrm{x}}^3-3{\mathrm{x}}^2-4\mathrm{x}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}{\mathrm{x}}^3-\frac{\mathrm{d}}{\mathrm{dx}}3{\mathrm{x}}^2-\frac{\mathrm{d}}{\mathrm{dx}}4\mathrm{x}\\ \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=3{\mathrm{x}}^2-3\cdot 2{\mathrm{x}}^{2-1}-4{\mathrm{x}}^{1-1}=3{\mathrm{x}}^2-6\mathrm{x}-4\dots ...\left(1\right)\end{array}$
Now the line $4\mathrm{x}+\mathrm{y}-3=0\Rightarrow \mathrm{y}=-4\mathrm{x}+3$
$\mathrm{y}=-4\mathrm{x}+3$ is the form of equation of a straight-line $\mathrm{y}=\mathrm{mx}+\mathrm{c}$, here m is the slope of the line.
Hence the slope $\mathrm{m}=-4=\frac{\mathrm{dy}}{\mathrm{dx}}$
Substitute this in equation (1)
$\begin{array}{l}\therefore -4=3{\mathrm{x}}^2-6\mathrm{x}-4\Rightarrow 3{\mathrm{x}}^2-6\mathrm{x}=0\Rightarrow 3\mathrm{x}(\mathrm{x}-2)=0\\ \Rightarrow 3\mathrm{x}=0\text{ or }\mathrm{x}-2=0\Rightarrow \mathrm{x}=0\text{ or }\mathrm{x}=2\end{array}$
For x=0 substitute it in $\mathrm{y}={\mathrm{x}}^3-3{\mathrm{x}}^2-4\mathrm{x}$
$\therefore \mathrm{y}=(0{)}^3-3(0{)}^2-4\left(0\right)=0$
Hence point is (0, 0)
For x=2 substitute it in $\mathrm{y}={\mathrm{x}}^3-3{\mathrm{x}}^2-4\mathrm{x}$
$\therefore \mathrm{y}=(2{)}^3-3(2{)}^2-4\left(2\right)=8-12-8=-12$
Hence point is (2, -12)
Therefore, points on the curve are (0, 0) and (2.-12)