Can $R^2$ be applied to non-linear least square regression? $R^2$ is usually used as a measure to determine a goodness of a fit. It appears to be used often times for linear least square fits, linear regression.
There's another measure which is RSS (residual sum of squares).
What's the difference from statistical application aspect?
Can $R^2$ be used for a non-linear least square fits/non-linear regression? if not why?

$R^2$ is a function of $RSS = \sum (y_i - \hat y_i)^2$.
$$
R^2 =
1-
\dfrac{
RSS
}
{
\sum (y_i - \bar y)^2
}
$$
What this equation means is that $R^2$ compares the model quality to always predicting the observed mean of $y$, regardless of your predictor variables.
Consequently, $R^2$ and $RSS$ provide the same information when it comes to model comparisons. If you compare two models of the same $y$, the one with lower $RSS$ has higher $R^2$, and this fact does not depend on model linearity.
I dislike $R^2$ for two reasons, both of which I have mentioned on here in the past, perhaps better than I describe here.
https://stats.stackexchange.com/a/539785/247274

*

*When the model is nonlinear, and even sometimes when it is linear, $R^2$ loses its interpretation from OLS of describing the proportion of explained variance.

Nonlinear regression SSE Loss
Why does regularization wreck orthogonality of predictions and residuals in linear regression?


*Even if we can interpret $R^2$ as the proportion of variance explained, people seem to want to think in terms of grades in school, where $R^2=0.9$ means an $\text{A}$-grade on our model, while $R^2=0.5$ means and $\text{F}$-grade on our model. Depending on the problem, $R^2=0.5$ might be fantastic performance that we would be very happy to achieve, while $R^2=0.9$ could be rather pedestrian performance.

Why getting very high values for MSE/MAE/MAPE when R2 score is very good
