Reciprocal solutions of a differential equation

You got so close. Suppose that $y_1=u$ is a solution, then $y_2=1/u$ must also be a solution. Substituting this, we obtain

$$ u'' -2\frac{(u')^2}{u} + \frac{b'}{b}u' + \frac{a^2}{b^2}u = 0 $$

But we already know that

$$ u'' + \frac{b'}{b}u' = \frac{a^2}{b^2}u $$

From the given assumption. Therefore

$$ 2\frac{a^2}{b^2}u - 2\frac{(u')^2}{u} = 0 \implies \frac{u'}{u} = \pm\frac{a}{b} $$

Integrating both sides gives

$$ u(x) = c\exp\left(\pm \int \frac{a}{b(x)} dx\right) $$

where $c$ is some arbitrary constant. Taking the positive and negative signs to be distinct solutions, we can put together the general solution

$$ y(x) = c_1 \exp\left(\int \frac{a}{b(x)}dx \right) + c_2 \exp\left(-\int \frac{a}{b(x)}dx\right) $$

which is indeed a linear combination of two reciprocal functions.

Hope this helps.