Question: I have a homework question that I don't know how to approach. Can I get some help?

Thanks

Find the points on $\displaystyle y = \frac{1}{2x-1}$ where the slope of the tangent line is $-2$.

Answer: 
The slope of the tangent line is given by the derivative $$y=(2x-1)^{-1}$$ Now applying the chain rule we get $$y'=-1(2x-1)^{-2}\cdot2=-\frac2{(2x-1)^2}$$We must set this value equal to $-2$ to find the desired points. $$\frac{-2}{(2x-1)^2}=-2\implies(2x-1)^2=1 \implies2x-1=\pm1$$ Thus $x=0,1$. Plugging this back into $y$ we get the points to be $$(0,-1)\ \text{and} \ (1,1)$$
Notice, the slope of the tangent at the curve $y=\frac{1}{2x-1}$ is given as $$\frac{dy}{dx}=\frac{-2}{(2x-1)^2}$$ $$\frac{-2}{(2x-1)^2}=-2$$
$$2x-1=\pm 1$$$$2x=1\pm 1\implies x=1, 0$$
setting the values of $x$, corresponding y-values are $y=1$ & $y=-1$ 

hence, the points are $\color{red}{\left(1, 1\right)}$ & $\color{red}{\left(0, -1\right)}$