Computation of the first few terms suggests the closed form is:
$$a_n=\frac{2}{n^2-n+2}$$
As a check we may write:
$$\frac{a_{n-1}}{1+(n-1)a_{n-1}}=\frac{\dfrac{2}{(n-1)^2-(n-1)+2}}{1+(n-1)\dfrac{2}{(n-1)^2-(n-1)+2}}=\frac{2}{(n-1)^2-(n-1)+2+2(n-1)}=\frac{2}{n^2-2n+1-n+1+2+2n-2}=\frac{2}{n^2-n+2}=a_n$$
This closed form satisfies the given initial value, and the given non-linear recurrence, and so we may state:
$$a_n=\frac{2}{n^2-n+2}$$