How to sketch the following discrete-time signal?
I give you some hints and then you can solve this homework.


Your $x[n]$ has only $8$ nonzero values. Figuring out what happens to them (in an exhaustive way) is not difficult.
Consider $(n-1)^2$ and change $n$ over $(-\infty, \infty)$. You can see that it is always non-negative. It means that we never can have $(n-1)^2=-1$. Hence the samples of $x[n]$ for $n=-3,-2,-1$ are never used. We can only have samples of $x[n]$ at $n=0,1,2,3$. 
Solve $(n-1)^2=0$, $(n-1)^2=1$,$(n-1)^2=2$, $(n-1)^2=3$. Only accept those results that lead to an integer $n$. Then you can see at what $n$ in $x[(n-1)^2]$ we have what value of $x[n]$. For instance, $(n-1)^2=\color{red}1\Rightarrow n=0,2$. It means in $x[(n-1)^2]$ the samples at $n=0$ and $n=2$ are equal to $x[\color{red}1]=1$.
Except at the points mentioned above, the transformed signal would be zero.