Finding $\frac{d^2 y}{dx^2}$ I am not sure how to do this but I need to find $\frac{dy}{dx}$ and $\frac{d^2 y}{dx^2}$
For $x = t^2 + 1, y= t^2+t$
And then show what t values gives a concave upward.
I know the simple formula to find $\frac{dy}{dx}$
I get $$\frac{dy}{dx} = \frac{y'}{x'}$$
$$\frac{dy}{dx} = \frac{2t+1}{2t}$$
$$\frac{d^2 y}{dx^2} = \frac{\frac{dy}{dx}}{dx}$$
$$\frac{\frac{2t+1}{2t}}{2t}$$
This is wrong and I am not sure why, they end with a negative number which makes no sense to me.

You have $$\frac{dy}{dx}=\frac{2t+1}{2t}=1+\frac1{2t}\;.$$ 
To differentiate this again with respect to $x$, you must repeat what you did to get this: calculate
$$\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{dx/dt}\;.$$
You forgot to do the differentiation in the numerator. When you do it, you get
$$\frac{\frac{d}{dt}\left(1+\frac1{2t}\right)}{2t}=\frac{-\frac1{2t^2}}{2t}=-\frac1{4t^3}\;.$$
