I have a confusion with something that should be very basic. Consider
$x = au(x,y)+bu(x,y)^2$
I want to take partial derivative with respect to y. I would think that the obvious form one gets is:
$ \frac{\partial x}{\partial y} = a u_y + 2b u(x, y) u_y$
However somehow, without thinking, I found myself writing down:
$ 0 = a u_y + 2b u(x, y) u_y$
I would think this is incorrect since we dont know what form $x(y)$ takes. But in my case $x$ and $y$ are independent random variables, would the LHS then be 0? I suppose I'm asking relatedly whether partial derivatives are causal. If y and x are created independently then one can not cause the other? Maybe that's why I wrote down the zero?
Your confusion is due to the bad notation that we use for partial derivatives. We write them as if they were fully defined by the variable with respect to which the derivative is taken. In fact, the definition of a partial derivative includes the specification of an entire set of variables that the function being differentiated depends on; but this is not apparent in the standard notation. If you write more explicitly
$$ \left.\frac\partial{\partial y}\right|_xx = \left.\frac\partial{\partial y}\right|_x\left(au(x,y)+bu(x,y)^2\right)\;, $$
where the variables under the vertical bar indicate the remaining independent variables that are being hold constant for purposes of the partial derivative, it’s clear that
$$ \left.\frac\partial{\partial y}\right|_xx=0\;. $$
Whether this is the partial derivative that you meant is not apparent from the notation you used.