Integrating by parts I get $$ \begin{align} \int_{0}^{1} \frac{\ln y}{(1-y)^2} \Big(\text{Li}_{2}(y) - \zeta(2) \Big) \ dy &= \int_{0}^{1} \frac{\ln^{2}(1-t)}{t} + \int_{0}^{1} \frac{\ln (t) \ln(1-t)}{1-t} \ dt \\ &= 2 \zeta(3) + \sum_{n=1}^{\infty} \frac{H_{n}}{(n+1)^{2}} \\ &= 2 \zeta(3) + \sum_{n=1}^{\infty} \frac{H_{n+1}}{(n+1)^{2}} - \sum_{n=1}^{\infty} \frac{1}{(n+1)^{3}} \\ &= 2 \zeta(3) + 2 \zeta(3) -1 - \zeta(3) +1 \\ &= 3 \zeta(3) \end{align}$$

You must have done the integration by parts differently.