Answer
$\dfrac{d\omega}{d\lambda}=-\dfrac{b^2\lambda}{a^2\omega}$
Work Step by Step
In order to derivate this function you have to apply implicit differentation method.
First, take the function to it's $f(\lambda)=0$ form
$ a^2\omega^2+b^2\lambda^2-1=0$
Then derivate the whole equation. Rember the put $\omega'$ every time you derivate $\omega$
$2a^2\omega \omega^2 + 2b^2\lambda=0$
Solve for a' and you have the answer
$\omega' (2a^2\omega)=-2b^2\lambda$
$\omega '=-\dfrac{b^2\lambda}{a^2\omega}$
