Partial derivative of differential equation Question
I have the following ordinary differential equation (ODE)
$\frac{du(s)}{ds}=f(s) + u(s)$
which can be written in my case as
$\frac{d}{ds}u[x(s),y(s)]=f[x(s),y(s)] + u[x(s),y(s)]$.
How do I find $\frac{\partial u}{\partial y}$?
What I tried
I take the partial derivative on both sides and I exchange the order of derivatives
$\frac{d}{ds}\left\{\frac{\partial}{\partial y}u[x(s),y(s)]\right\}=\frac{\partial}{\partial y}\left\{f[x(s),y(s)] + u[x(s),y(s)]\right\}$
now, I define
$g(s)=\frac{\partial u}{\partial y}$
which implies I can find $\frac{\partial u}{\partial y}$ by solving the following ODE for $g(s)$
$\frac{dg(s)}{ds}=\frac{\partial}{\partial y}f[x(s),y(s)] + g(s)$
Could someone confirm whether this is correct?
Counter-Example
Here is an example where my above statement is wrong. I define
$x(s)=\cos(s)\\ 
y(s)=s^2\\
u(x,y)=2x+y^2\\
f(x,y)=4y^{3/2}-y^2-2x-\sqrt{1-x^2}$
one can verify with some algebra that it satisfies (by design)
$\frac{du(s)}{ds}=f(s) + u(s)$
in fact, both sides are equal
$\frac{du(s)}{ds}=4s^3-2\sin(s)=f(s) + u(s)$
However
$\frac{d}{ds}\left\{\frac{\partial}{\partial y}u[x(s),y(s)]\right\}=4s$
while
$\frac{\partial}{\partial y}\left\{f[x(s),y(s)] + u[x(s),y(s)]\right\}=6s$
So my original question remains: given an ODE for $u(s)$ what ODE defines $\frac{\partial u}{\partial y}$?
Solution
Based on feedback from @lord-commander, I realized that my proposed approach was completely wrong. In fact, the unknown partial derivative $\partial u/\partial y$ that I am looking for is related to the given functions by the following partial differential equation, not by an ODE.
$\frac{\partial u}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial s}=f(s)+u(s)$
It seems to me then that my only option is to solve the original linear ODE
$\frac{du(s)}{ds}=f(s) + u(s)$
which has a general solution
$u(s) = c_1 e^s + e^s \int_1^s f(s') e^{-s'} ds'$
Then I write this as a function of $(x,y)$ and take the partial derivative.
This is the approach I was trying to avoid. In fact, my expression is complicated and the integral has to be done numerically, for many values of $s$ with $f(s)$ a computationally expensive function.
 A: I have the following differential equation
$$\frac{du(t)}{dt}=f(t) + u(t)$$

I decided to use $t$ because my eyes just didn't work with $s$, but feel free to read the last equation and change the $t$-s back to $s$.

which can be written in my case as
$$\frac{d}{dt}u(x,y)=f(x,y)+ u(x,y)$$
Both $x(t)$ and $y(t)$ depend parametrically on $t$.
$$u(x(t),y(t))=u(x,y)$$
By the chain rule:
$$\frac{du}{dt}=\frac{d}{dt}u(x,y)=\frac{\partial u}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial t}$$
$d$ is the total derivative, $\partial$ is a partial derivative.
$$\frac{\partial u}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial t}=f(x,y)+u(x,y)$$
$$\frac{d}{dy}\left(\frac{\partial u}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial t}\right)=\frac{d}{dy}\left(f(x,y)+u(x,y)\right)$$
$$\frac{d\;\frac{\partial u}{\partial x}\frac{\partial x}{\partial t}}{d y}+\frac{d\;\frac{\partial u}{\partial y}\frac{\partial y}{\partial t}}{dy}=\frac{d}{dy}\left(f(x,y)+u(x,y)\right)$$
$$\frac{\partial}{\partial t}\left(\frac{\partial u}{\partial x}\frac{\partial x}{\partial t}\right)\cdot \frac{\partial t}{\partial y}+\frac{\partial}{\partial t}\left(\frac{\partial u}{\partial y}\frac{\partial y}{\partial t}\right)\cdot\frac{\partial{t}}{\partial y}=\frac{d}{dy}\left(f(x,y)+u(x,y)\right)$$
$$\frac{\partial}{\partial t}\left(\frac{\partial u}{\partial x}\frac{\partial x}{\partial t}\right)\cdot \frac{\partial t}{\partial y}+\frac{\partial}{\partial t}\left(\frac{\partial u}{\partial y}\frac{\partial y}{\partial t}\right)\cdot\frac{\partial{t}}{\partial y}=\left(\frac{\partial f}{\partial x}\frac{\partial x}{\partial y}+\frac{\partial u}{\partial y}\underbrace{\frac{\partial y}{\partial y}}_{1}\right)$$
To be completely frank i am not really sure if you can multiply out by $\frac{\partial }{\partial t}$ as the functions they are differentiating are different. I could be wrong though.
