I am attempting to differentiate this function, but I am not having success in getting rid of fractions so that I can separate $\frac{dy}{dx}$ onto one side of the question. These are the steps I have followed:
Step 1
$$2y\frac{dy}{dx}-\frac{(x-1)\frac{dy}{dx}-y}{(x-1)^2}=0$$
Step 2
$$\frac{(x-1)^2(2y)\frac{dy}{dx}-(x-1)\frac{dy}{dx}-y}{(x-1)^2}=0$$
I am unsure of where to go from here. Any advice would be greatly appreciated.
$$(x-1)^2(2y)\frac{dy}{dx}-(x-1)\frac{dy}{dx}-y=0$$ $$(x-1)^2(2y)\frac{dy}{dx}-(x-1)\frac{dy}{dx}=y$$ $$\left[2y(x-1)^2-x+1\right]\frac{dy}{dx}=y$$ $$\frac{dy}{dx}=\frac{y}{2y(x-1)^2-x+1}$$
You are on the right track. Try writing it like this:
$$ \begin{aligned} &\frac{(x-1)^{2}(2y)\frac{dy}{dx}-(x-1)\frac{dy}{dx}}{(x-1)^{2}}-\frac{y}{(x-1)^{2}}=0\\[5pt] \implies&\frac{(x-1)^{2}(2y)\frac{dy}{dx}-(x-1)\frac{dy}{dx}}{(x-1)^{2}}=\frac{y}{(x-1)^{2}}\\[5pt] \implies &\frac{dy}{dx}\frac{(x-1)^{2}(2y)-(x-1)}{(x-1)^{2}}=\frac{y}{(x-1)^{2}}\\[5pt] \implies&\frac{dy}{dx}\frac{(x-1)(2y)-1}{x-1}=\frac{y}{(x-1)^{2}}\\[5pt] \implies &\frac{dy}{dx}=\frac{y}{(x-1)^{2}}\cdot\frac{x-1}{(x-1)2y-1}\\[5pt] \implies &\frac{dy}{dx}=\frac{y}{(x-1)^2(2y)-(x-1)}. \end{aligned} $$
You could also use implicit differentiation considering $$F=y^2 - \frac{y}{x-1}-4=0$$ So $$F'_x=\frac{y}{(x-1)^2} \qquad , \qquad F'_y=2 y-\frac{1}{x-1}$$ $$\frac{dy} {dx}=-\frac{F'_x }{F'_y}=???$$
