<s>[INST] a line tangent to curve $y = \frac{x}{2−2x}$ goes through (1, -1) $y = \frac{x}{2−2x}$
$y = \frac{f(x)}{g(x)} \rightarrow y' = \frac{f'(x) \cdot g(x) - g'(x)\cdot f(x)}{{g(x)}^2}$
$y = \frac{x}{2−2x}$
$y' = \frac{(2-2x)-(x \cdot (-2))}{{(2−2x)}^2}$ $\rightarrow$ $y' = \frac{(2-2x+2x)}{{(2−2x)}^2} = \frac{2}{{(2-2x)}^2}$
why cant i find the gradient with derivative?
 [/INST] Your derivative is absolutely correct.
Equation of a line is $(y-y_1)=m(x-x_1)$ where $(x_1,y_1) $ is a point that lies on the line
Here $m = \frac{2}{(2-2x)^2}$
Now we know that $(1,-1)$ lies on the line and the gradient of the line is $\frac{2}{(2-2k)^2}$
Substitute everything
We get $y=\frac{2}{(2-2k)^2}(x-1)-1$
Now this line also passes from point and $(k,\frac{k}{2-2k})$
We get $\frac{k}{2-2k}=\frac{2}{(2-2k)^2}(k-1)-1$ 
Now solve this equation for k and you'll get $k=3$
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