The standard techniques is as follows. At first, we guess the growth rate by replacing the discrete dynamical system to ODE: $a_n-a_{n-1}=-a_{n-1}\phi(a_{n-1})$ is replaced to $A'(n)=-A(n)\phi(A(n))$, where function $A(n)$ imitates $a_n$. Nex, we solve this equation, rewriting it as $g(A(n))'=1$, where $g'(t)=-1/(t\phi(t))$. Our guess is that $g(A(n))$ behaves like $n$ (since it derivative behaves as 1), and the same for $g(a_n)$. For proving this guess rigorously we introduce a new sequence $b_n=g(a_n)$ and write down the formula $b_n-b_{n-1}=g(a_n)-g(a_{n-1})=(a_n-a_{n-1})g'(s)$ for some $s$ between $a_{n-1}$ and $a_n$. Next, $(a_n-a_{n-1})g'(s)=\frac{a_{n-1}\phi(a_{n-1})}{s\phi(s)}$. Assume that $\phi(x):\phi(y)$ tends to 1 when $x/y$ tends to 1 and $x,y$ tend to $+0$. Then we may conclude that our last fraction tends to $1$, and by usual Stolz theorem conclude that $b_n$ behaves like $n$.

Example: $\phi(x)=x$, $g(t)=1/t$, so we consider the sequence $b_n=1/a_n$. We get $b_{n}-b_{n-1}=\frac{a_{n-1}-a_{n}}{a_na_{n-1}}=\frac{a_{n-1}}{a_{n}}$, what clearly tends to 1, hence $b_n\sim n$.