This could be a silly question. In my lecture note on the Lipschitz condition, there is an example, which is
$y(0)=1$ and $f(t,y)=y-t^2+1$, determine whether this IVP is well-posed.
Here we apply the Lipschitz condition, namely $|\frac{\partial f}{\partial y}|=1$. But I think $\frac{\partial f}{\partial y}=\frac{\partial f}{\partial t}\frac{\partial t}{\partial y}+\frac{\partial f}{\partial y}=-2t\frac{\partial t}{\partial y}+1$, which generally contradicts the result above, since $y$ is a function of $t$. How to explain this?
I have considered the condition that $t$ is fixed, which makes $\frac{\partial t}{\partial y}=0$. But then another contradiction pops up, if $t$ is fixed, then $f, y$ should also be constants, why is $|\frac{\partial f}{\partial y}|\neq0$?