In the attached problem there are two parts I had to figure out. For part a) I had to find dy/dx in terms of the variable t using the information stated in the top. However, I'm not confident about my answer for part b). Can anyone check to see that I have answered part b) correctly? My answer for part b) is at the bottom right of the image. Thank you!
On
The answer for a) if you need it is simple. $\dfrac{dy}{dx}=\dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}.$ Substitute $\dfrac{dy}{dt} = 2t$ and $\dfrac{dx}{dt}= \dfrac{1}{(1+t)^2}.$
So we have $\dfrac{dy}{dx} = 2t(1+t)^2$ and $x=1-\dfrac{1}{1+t}.$ Then $1+t = \dfrac{1}{1-x}$ and $t=\dfrac{1}{1-x}-1.$ Just substitute the values into the expression for $\dfrac{dy}{dx}$ to get your answer.
Your work is correct except for a minus sign. In your final answer, the denominator is $x^3-3x^2+3x\mathbf{-}1$, not $x^3-3x^2+3x\mathbf{+}1$? The correct answer has a minus sign. I'll just note there's a slightly easier way. $$y=\frac{x^2}{x^2-2x+1}=\left(\frac{x}{x-1}\right)^2$$ so $$y'= 2\frac{x}{x-1}\left(\frac{x}{x-1}\right)'$$