Suppose we have $\,y(t) = x(-t+1)\,.$
I want to prove that $\,y(t)\,$ is time-invariant.
Reading from wikipedia
More generally the relationship between the input and the output is $\,y(t) = f\big(x(t),t\big)$ ,
and its variation with time is
$\dfrac{dy}{dt}=\dfrac{\partial f}{\partial t}+\dfrac{\partial f}{\partial x}\dfrac{dx}{dt}$
For time invariant systems, the system properties remain constant with time, $\dfrac{\partial f}{\partial t}=0$.
In my case $\dfrac{\partial f}{\partial t}=0$ is true so the system should be time invariant. Is this correct ? And if so, can I use the above to always check if a system is time-invariant?
The simplest test for time invariance is that if
$$ x(t) \Rightarrow y(t)$$
then $$ x(t+d) \Rightarrow y(t+d) $$
Which simply means "a given time shift of the input signal results in the same time shift of the output signal".
So given the system
$$ y(t) = x(-t+1) $$
we want to check if shifting the input by some arbitrary amount $d$ results in the output being shifted by that same amount $d$: $$y(t+d)\stackrel{?}{=} x(-t+1+d)$$
So working through the algebra we find:
$$\begin{align*}x(\left[-t+1\right]+d) &= x(-[t-d]+1)\\ &= y(t-d)\\ &\ne y(t+d)\\ \end{align*}$$
and we can conclude the system is not time-invariant.
In fact shifting the input in one direction, shifts the output in the opposite direction. This system is very dependent on the timing of the input relative to time $t=0$, since the system reverses the time axis in the output.