How do I simplify $\frac{(bS)^{2}}{(bS)^{2} + y}$ to $\frac{1}{1+\left[\frac{y}{bS}\right]^{2}}$?

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As the above mentions I have the fraction $\frac{(bS)^{2}}{(bS)^{2} + y}$ and the next step in the equation I am following simply states "it works out to equal" $\frac{1}{1+\left[\frac{y}{bS}\right]^{2}}$.

I'm sure its simple but I am very rusty on my algebra, do I need to multiply by the conjugate?

I'd be very grateful for a worked example

Thanks in advance

edit: I accidentally wrote - when I meant to put + in both denominators. Below is a clarification of terms.

The initial expression was $\frac{M^{2}}{M^{2}+y}$, where M = bS

edit2: Having checked the paper the text I'm following is based on I have found that the initial expression should have been $\frac{M^{2}}{M^{2}+y^{2}}$, solving the issues I was having.

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As the good people in the comments suspected, it turns out that the reason I was struggling was because of a typo in the text I was following.

$\frac{(bS)^{2}}{(bS)^{2} + y}$ should have been $\frac{(bS)^{2}}{(bS)^{2} + y^{2}}$

Simplification is therefore a process of dividing both the numerator and denominator by $(bS)^{2}$ to end up with $\frac{1}{1 + [\frac{y}{bS}]^{2}}$