In Euler-Lagrange equations, why we take ${\partial T}/{\partial {x}} $ as zero (when no terms of $x$ is present)? Basically, why we treat them as independent quantities. I know what a partial derivative is, It means if a function depends on multiple variables, the partial derivative with respect to a particular variable will show how the function is varying with that variable while other variables are constant.
Now, we know $x=f(t)$ and $v=f'(t). $
So we know $v=dx/dt$.
So how is $ {\partial T}/{\partial {x}} =0 ? $ ($T$=Kinetic energy)
Because this would be true only for those values of $x$ where the slope of  $v - x $ curve is zero.
For instance, if $x=t^3$ , so $v=3t^2$ or $v=3x/t$.  Here, ${\partial T}/{\partial {x}} =0$ is clearly not zero.
Basically I want to ask why do we take x and v independent of each other in variational calculus. Because we do know that they are dependent quantities.
 A: *

*The Lagrangian $L$ is a function $(q,v,t)\mapsto L(q,v,t)$ where its arguments $(q,v,t)$ are independent. To form the partial derivatives of $L$ wrt. the variables $(q,v,t)$ it is necessary that the arguments $(q,v,t)$ are independent.
Above it is implicitly assumed that we are not using the EOMs. If we are using the EOMs, then the variables $(q,v,t)$ are clearly no longer independent. This seems to be the core of OP's problem.
See also this related Phys.SE post.


*For a concrete example where $T$ depends on generalized position coordinates, cf. OP's title question, think of a free particle in spherical coordinates.
A: If $T = \frac 1 2 m v^2 = \frac 1 2 m (\dot x)^2$ then $T$ has no explicit dependence on $x$, so $\frac {\partial T}{\partial x}=0$.
Note that this does not mean that $T$ has no implicit dependence on $x$, neither does it imply that $\frac {dT}{dx} = 0$. In your example,
$x = t^3 \\ \Rightarrow \dot x = 3t^2 = 3 x^{\frac 2 3}
\\ \displaystyle \Rightarrow \frac {dT}{dx} = \frac {\partial T}{\partial x} + \frac {\partial T}{\partial \dot x} \frac {d \dot x}{ dx} = 0 + (m \dot x) (2 x ^{- \frac 1 3}) = 6 m x^{\frac 1 3} = 6mt$
A: It depends on the context. In physics the implicit dependence of functions is often let out which makes a lot of expressions easier to read but in cases like this it can get confusing. When $\partial T/\partial x=0$ was written the author was probably considering this:
$$H(x,v)=T(v)+U(x)$$
Here $H$ is just a function of two variables. Nothing more. When you consider a path $x(t)$ it is useful to put $v=\frac {dx}{dt}$. So you calculate
$$H(x,\frac{dx}{dt})=T(\frac{dx}{dt})+U(x)$$
In this case calculating derivatives is considerably more confusing but remember that we started from the first equation and the author implicitly means 'take the derivative with respect to the second variable, which I call $v$'.
The proces you describe can be summarised as follows
$$v(x)=v(t(x))\\
\frac{\partial T}{\partial x}=\frac{\partial T}{\partial v}\frac{\partial v}{\partial t}\frac{\partial t}{\partial x}$$
where $t(x)$ is a function that is obtained by inverting $x(t)$.
