On Exercise 2.5.10 in Ram. M. Murty's book, "Problems in Analytic Number Theory."

It seems there is a typo in the application of the hyperbola method. Since $b_n=\sum_{d\mid n}a_d$, we have \begin{align} \sum_{n\le x}b_n &=\sum_{n\le x}\sum_{de=n}a_d=\sum_{de\le x}a_d\\ &=\sum_{\substack{de\le x\\ d\le y}}a_d+\sum_{\substack{de\le x\\ e\le x/y}}a_d-\sum_{de\le x\\ d\le y, e\le x/y}a_d. \end{align} If $A(x)=\sum_{n\le x}a_n$, then this can be written as \begin{align} \sum_{d\le y}a_d\left[\frac{x}{d}\right]+\sum_{e\le x/y}A\left(\frac{x}{e}\right)-A(y)\left[\frac{x}{y}\right]. \end{align} The author's assumption that $A(x)\ll x^{\delta}$ then gives the last two terms as $O(xy^{\delta-1})$, and the proof can be continued in the usual manner outlined in the text. Although, it appears that a sign was missed later on as well, which has some bearing on the final stated error (see Conrad's answer below).

The computation of the powers is wrong and the result stated in the book is incorrect and it should be $cx+ O(x^{(1+\delta-\delta^2)/(2-\delta)})+O(x^{\frac{(1-\delta)(1+\alpha)}{2-\delta}})$

If $y=x^{(1-\delta)/(2-\delta)}$, $y^{\delta-1}=x^{-(1-\delta)^2/(2-\delta)}$, so $xy^{\delta-1}=x^{(1+\delta-\delta^2)/(2-\delta)}$ and it is not true that $\frac{1+\delta-\delta^2}{2-\delta} \le \frac{(1-\delta)(1+\alpha)}{2-\delta}$ in general, only for $\alpha \ge \delta + \frac{\delta}{1-\delta}$

Note that as stated, the result doesn't make sense because one can always increase $\delta$ in the hypothesis, while keeping $\alpha$ fixed, so in particular if $A(x) << x^{\delta}$ for a given $\delta < 1$, then $A(x) <<_{\epsilon} x^{1-\epsilon}$ for arbitrary $\epsilon >0$, hence we would get that $B(x) =cx + O_\epsilon(x^{\epsilon})$ under very general conditions and I am sure lots of counterxamples to that can be found