Find the coordinates of any stationary points on the curve $y = {1 \over {1 + {x^2}}}$ and state it's nature I know I could use the quotient rule and determine the second differential and check if it's a max/min point, the problem is the book hasn't covered the quotient rule yet and this section of the book concerns exercises related only to the chain rule, so I was wondering what other method is there to determine the nature of the stationary point (which happens to be (0,1)) given that this is an exercise that is meant to utilise the chain rule.

$${{dy} \over {dx}} =  - {{2x} \over {{{({x^2} + 1)}^2}}}$$
So essentially my question is can I determine ${{{d^2}y} \over {d{x^2}}}$ by using the chain rule (which I dont think you can) and thus the nature of stationary point, or would I have to determine the nature of the stationary point another way?
I think I may have overlooked something, any help would be appreciated. Thank you.

I'm sorry, but it's worth noting that $$\frac{d^2y}{dx^2}=\frac{6x^2-2}{\left(x^2+1\right)^3},$$ not what has been previously posted. You can't actually use the chain rule for this, unless there's a very, very clever trick.
You're going to have to find the nature of the stationary points by a different method - the first-derivative test, rather than the second-derivative test.
This is a fancy name for "is it increasing/decreasing just to the left?" and "is it increasing/decreasing just to the right?" You know that the only stationary point is 0, since that's the only way $\frac{dy}{dx}$ can be 0. To the left of 0 (i.e. with $x$ small and negative), is $\frac{dy}{dx}=\frac{-2x}{\left(x^2+1\right)}$ positive or negative? The denominator is always positive... the numerator is positive when $x$ is negative, so $\frac{dy}{dx}>0$ for $x<0$. In other words, $y$ is increasing for negative $x$.
Similar reasoning will show you that $\frac{dy}{dx}<0$ for $x>0$, so $y$ is decreasing for positive $x$.
If $y$ is increasing to the left of 0, and decreasing to the right of 0, then 0 must be a local maximum.
