Why is the partial derivative w.r.t. a derivative legitimate? Gilbert Strang linearize two ODEs in this video. I don't follow the steps there he takes the partial derivative w.r.t. to the function $y(t)$ and the derivative $y'(t)$ (same for $z(t),z'(t)$). No chain rule is used. 
He takes the partial derivatives and treat the functions as constants (!).
Why is it correct? Is it a legitimate way?
From the video (I don't supress the arugments of the functions):

The critical point is $(y(t),z(t))=(Y,Z)=(0,0)$, where I guess $Y,Z$ are constants.
  \begin{align}
\frac{dy}{dt}&=y(t)-y(t)z(t) =f(y(t),z(t))\\
\frac{dz}{dt}&=y(t)z(t)-z(t) =g(y(t),z(t))
\end{align}
  And the Taylor expansion
  \begin{align}
f(y(t),z(t))&=f(Y,Z)+\frac{\partial f}{\partial y(t)}(y(t)-Y)
+\frac{\partial f}{\partial z(t)}(z(t)-Z)\\
g(y(t),z(t))&=g(Y,Z)+\frac{\partial g}{\partial y(t)}(y(t)-Y)
+\frac{\partial g}{\partial z(t)}(z(t)-Z)\\
\end{align}
  So $\frac{\partial f}{\partial y(t)}=1-z(t)$, $\frac{\partial f}{\partial z(t)}=-y(t)$, $\frac{\partial g}{\partial y(t)}=z(t)$ and $\frac{\partial g}{\partial z(t)}=y(t)-1$. 
But hey, why no chain rule here? Isn't $\frac{\partial f}{\partial y(t)}=y(t)y'(t)-y'(t)z(t)+z'(t)y(t)$? And so on for the rest.

 A: The key to why the chain rule is not applied is that in the linearization of f and g the derivative is taken with respect to y and z, not t. If we take
$\frac{\partial f}{\partial y(t)}= 1-z(t)$
but if we take
$\frac{\partial f}{\partial t} = y'(t)(1-z(t)) + y(t)z'(t)$
as you suspected.
A: You need to separate two situations or contexts. The first is the definition, or construction, of the functions $f,g$ and their linearization via Taylor series. There $f,g$ are just a functions in two variables $y,z$ and all partial derivatives in the Taylor expansions are just that, ordinary partial derivatives $f_y=\dfrac{∂f}{∂y}(Y,Z)$ and $f_z=\dfrac{∂f}{∂z}(Y,Z)$, and similar for $g$, with the linearization then 
$$
f(y,z)\approx f(Y,Z)+f_y\,(y-Y)+f_z\,(z-Z),\\
g(y,z)\approx g(Y,Z)+g_y\,(y-Y)+g_z\,(z-Z).
$$
The second context is the use of these two functions as right side in a differential equation where then indeed one has function values as arguments
$$
\dot y(t)=f(y(t),z(t)),\\
\dot z(t)=g(y(t),z(t)).
$$
The claim used then is that if $(y(0),z(0))$ are close to $(Y,Z)$, then for some time the solutions $(y(t),z(t))$ of the original equation and $(\bar y(t),\bar z(t))$ of the linearized equations
$$
\dot {\bar y}(t)=f(Y,Z)+f_y\,(\bar y(t)-Y)+f_z\,(\bar z(t)-Z),\\
\dot {\bar z}(t)=f(Y,Z)+g_y\,(\bar y(t)-Y)+g_z\,(\bar z(t)-Z)\\
$$
stay close together. This is especially true for a stable stationary point, but in general any stationary point can be analyzed that way, one only has to make sure that "close" is as close as necessary. Then the leading terms are the linear terms and they dominate all other terms in the Taylor expansion.
