find a horizontal speed $\frac{dx}{dt}$ A girl slides down a slide in the shape of the parabola $y=(x−1)^2$ for $0≤x≤1$. Her vertical speed is $\frac{dy}{dt}=−y(1−y)$. Find her horizontal speed $\frac{dx}{dt}$ when $y=\frac 12$.
I have found that $$\frac{dy}{dt}=2(x+1)\frac{dx}{dt}$$, since $\frac{dy}{dt}=−y(1−y)$ and $y=\frac 12$ I got $$\frac{dx}{dt}=\frac {-\frac 12(1-\frac 12)}{2(x+1)}$$, but what to do with x now?

Use the chain rule:
$$
\frac { dx }{ dt } \quad =\quad \frac { dx }{ dy } \frac { dy }{ dt } \\ \qquad \quad =\quad \frac { 1 }{ \frac { dy }{ dx }  } y(y-1)\\ \qquad \quad =\quad \frac { y(y-1) }{ 2(x-1) } \\ \frac { dx }{ dt } \quad =\quad \frac { y(y-1) }{ -2\sqrt { y }  } 
$$
And now all you have to do is substitute $y = \frac{1}{2}$ to get:
$$
\\ \frac { dx }{ dt } \quad =\quad \frac { \frac { 1 }{ 2 } (\frac { 1 }{ 2 } -1) }{ -2\sqrt { \frac { 1 }{ 2 }  }  } \quad =\quad \frac { 1 }{ 4\sqrt { 2 }  } 
$$
