I have the function $y(t)=t^2x(t-1)$ and I need to figure out if it is linear or not and time invariant or not.
By the looks of it I guessed it to be not linear but the answer is linear but not time invariant. I can't think of a way to prove it.
Can we always say that a function is time invariant if the coefficient of $x(t)$ (the input) does not contain any $t$'s?
You can answer such questions in a very systematic way. For linearity, you need to show that if $y_1(t)$ is the response to $x_1(t)$, and if $y_2(t)$ is the response to $x_2(t)$, the response to $x(t)=ax_1(t)+bx_2(t)$ must equal $y(t)=ay_1(t)+by_2(t)$. For your system we have
$$y_1(t)=t^2x_1(t-1)\quad\text{and}\quad y_2(t)=t^2x_2(t-1)$$
and
$$y(t)=t^2x(t-1)=t^2(ax_1(t-1)+bx_2(t-1))=ay_1(t)+by_2(t)$$
Hence, the system is linear.
For time-invariance you need to show that if $y(t)$ is the response to $x(t)$, then the response to $x(t-T)$ must equal $y(t-T)$ for any shift $T$. For $y(t)=t^2x(t-1)$ we get $y(t-T)=(t-T)^2x(t-T-1)$. However, the response to $x(t-T)$ is given by
$$t^2x(t-T-1)\neq y(t-T)$$
Consequently, the system is NOT time-invariant, but it is time-varying.