Saving increasing in interest rate
$$y = \frac {a(x)}{b+a(x)} \implies \frac{\partial y}{\partial x} = \frac{\partial y}{\partial a}\cdot \frac{\partial a}{\partial x}$$

Since

$$\frac{\partial y}{\partial a} = \frac {b+a -a}{(b+a)^2} = \frac {b}{(b+a)^2} &gt;0$$

because in your case "$b$"$&gt;0$,

the sign of $\frac{\partial y}{\partial x}$ will be the same as the sign of $\frac{\partial a}{\partial x}$, which is what Romer is asserting.

$\theta &gt;1$ would make $\frac{\partial a}{\partial x} &lt;0$, so 

$$\theta &lt;1 \implies \frac{\partial a}{\partial x}&gt;0\implies \frac{\partial y}{\partial x}&gt;0$$