The slope of the tangent at $(-1,f(-1))$ can be found by differentiating $f(x)$ and plugging in $x=-1$. Based on your work, it looks like want to evaluate the slope using first principles. Hence, slope at $x=-1$ is given by \begin{align} \lim_{h \to 0} \dfrac{f(-1+h)-f(-1)}h & = \lim_{h \to 0} \dfrac{\left(\dfrac{-1+h+2}{1-(-1+h)}-3(-1+h) \right) - \left(\dfrac{-1+2}{1-(-1)}-3(-1) \right)}h\\ & = \lim_{h \to 0} \dfrac{\left(\dfrac{h+1}{2-h}+3 - 3h \right) - \left(\dfrac{1}{2}+3 \right)}h = \lim_{h \to 0} \dfrac{\left(\dfrac{h+1}{2-h} - 3h \right) - \dfrac{1}{2}}h\\ & = \lim_{h \to 0} \dfrac{\overbrace{\left(\dfrac{h+1}{2-h} - \dfrac{1}{2} \right)}^{\text{Take the l.c.m}} - 3h}h\\ & = \lim_{h \to 0} \dfrac{\left(\dfrac{2h+2-2+h}{4-2h} \right)-3h}h = \lim_{h \to 0} \dfrac{\left(\dfrac{3h}{4-2h} \right)-3h}h\\ & = \lim_{h \to 0} \left(\dfrac3{4-2h} - 3 \right) = \dfrac34 -3 = -\dfrac94 \end{align} Hence, the slope of the tangent at $(-1,f(-1))$ is $-9/4$. You also know that the tangent passes through $(-1,f(-1))$. Note that $f(-1) = \left(\dfrac{-1+2}{1-(-1)}-3(-1) \right) = \dfrac72$ Hence, the tangent passes through $(-1,7/2)$ and has slope $-9/4$. The equation of the tangent is $\dfrac{y-7/2}{x+1} = \dfrac{-9}4 \implies y = \dfrac72 - \dfrac94(x+1) = -\dfrac94x + \dfrac54$ Hence, the equation is $4y+9x=5$