Prove that $ y''(0) = -1, x = \cos\left(\frac{t}{1+t}\right), y = \sin\left(\frac{t}{1+t}\right)$ My attempt:

$$x'= -\sin\left(\frac{t}{1+t}\right) * \frac{1}{(1+t)^2}$$

$$y'= \cos\left(\frac{t}{1+t}\right) * \frac{1}{(1+t)^2}$$

then I attempted to divide y' on x' which resulted in

$$-\cot\left(\frac{t}{1+t}\right)$$

When I try to take the double dash of this it gives me 

$$\csc^2\left(\frac{t}{1+t}\right),$$

but this function is undefined at 0.
The second derivative of a parametric curves is
\begin{align*} 
&amp;\frac{d}{dt} \left( \frac{dy}{dx} \right) / \left( \frac{dx}{dt} \right)\\
=&amp;\frac{d}{dt} \left( -\cot \left( \frac{t}{1+t} \right) \right) / \left( \frac{d}{dt} \left( \sin\left( \frac{t}{1+t} \right) \right) \right)\\
=&amp;\frac{\csc^2\left( \frac{t}{1+t} \right)}{\left( 1+t \right)^2} \cdot \frac{\left( 1+t \right)^2}{-\sin\left( \frac{t}{1+t} \right)}\\
=&amp;-\csc^3\left( \frac{t}{1+t} \right)
\end{align*}
This is not defined for $t=0$. 

You can see this easily by $\frac{t}{1+t}\mapsto t$. Now you can more clearly see that it traces the unit circle with $t=0$ giving the point $(0,1)$. Intuitively, the slope is infinitely great at this point.
To prove y''(0)=-1

Take the derivative of the second equation using the product, quotient and chain rules. Then replace $t$ with $0$ and evaluate.

For the other two you can integrate:

$\int -\sin(\frac{t}{1+t})*\frac{1}{(1+t)^2}*dt$

$u=\frac{t}{1+t}$

$du=\frac{1}{(1+t)^2}*dt$

$\int -\sin(u)*du$

Which I think you can evaluate :)