Time variant and Time Invariant Systems
Here is a simple way of looking at it which avoids any algebra and uses minimal symbolism. Let us calculate the output $y$ for an input $x$ which is a unit pulse function. We have that


$y[-2] = (-2)(x[-2])^2 = 0$ since $x[-2] = 0$.
$y[-1] = (-1)(x[-1])^2 = 0$ since $x[-1] = 0$.
$y[0] = (0)(x[0])^2 = 0$ (regardless of the value of $x[0]$).
$y[1] = (1)(x[1])^2 = 0$ since $x[1] = 0$.
$y[2] = (2)(x[2])^2 = 0$ since $x[2] = 0$.


and so on.  The entire input is blocked and the output $y$ is always $0$.

What about the output $\hat{y}$ when the input $\hat{x}$ is just the previous $x$
delayed by one time unit? The only difference now is that $\hat{x}[1] = 1$ and
we have that 


$\hat{y}[1] = (1)(\hat{x}[1])^2 = 1$ since $\hat{x}[1] = 1$.


It is, I hope, obvious that $\hat{x}[n] = 0$ for all integers $n$ except $n=1$.
So, since the input $\hat{x}$ is just a delayed version of input $x$, is
is true that the output $\hat{y}$ is just a delayed version of output $y$?
What, if anything, does this tell you about the system properties?