As you know already, derivatives measure how a quantity varies with respect to another variable. Actually, partial derivatives take into account direct contributions only, while total derivatives include indirect contributions too through the chain rule.
Let's illustrate this matter of fact by taking a bivariate function $f(x,y)$ as an example. Its differential, which corresponds to an infinitesimal variation of $f$, is given by
$$
\mathrm{d}f(x,y) = \frac{\partial f}{\partial x}\mathrm{d}x + \frac{\partial f}{\partial y}\mathrm{d}y.
$$
Here, $\frac{\partial f}{\partial x}$ tells you how much the function $f$ varies when its first variable evolves from $x$ to $x + \mathrm{d}x$ independently of $y$, and the same goes for the second term mutatis mutandis. Now, let's suppose you want to compute the total derivative of $f$ with respect to $x$, i.e.
$$
\frac{\mathrm{d}f}{\mathrm{d}x} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}x}.
$$
As you can observe, the contributions to the said total derivative consist in the direct contribution of $x$ through the associated partial derivative, as well as the indirect contribution of $y$ as a function of $x$ through the chain rule $-$ the factor $\frac{\partial f}{\partial y}$ being the direct contribution to the variation of $f$ by $y$ and $\frac{\mathrm{d}y}{\mathrm{d}x}$ the one of $y$ by $x$.
In conclusion, since partial derivatives measure "direct" variation rates only, all the other variables have to be considered as constant parameters.
Addendum.
Since I haven't subscribed to MathOverflow, I will leave here a few words about your concerns discussed originally in this post.
Firstly, note that the functional derivative is the infinite-dimensional analog of the total derivative. In consequence, as you foresaw, in the case of a functional depending on the inverse function, namely
$$
S[x] = \int L(t,x(t),x^{-1}(t)) \,\mathrm{d}t
$$
in the present case, the contribution of $x^{-1}(t)$ to the functional differential of $S[x]$ cannot be ignored. However, analogously to the classical case, the said contribution doesn't originate from the partial derivative of $L$ with respect to $x$, but it will come from the chain rule and the derivative with respect to $x^{-1}$ itself. More concretely, one has :
$$
\delta S[x] = \int \left(\frac{\partial L}{\partial x}\delta x(t) + \frac{\partial L}{\partial x^{-1}}\delta x^{-1}(t)\right)\mathrm{d}t
$$
Now, the inverse function function may be reinterpreted as the $(-1)^\mathrm{th}$ iterate. You may prove (see here for a brief introduction) that the functional derivative of a function composed $n$-times with itself is given by
$$
\frac{\delta x^{\circ n}(t)}{\delta x(\tau)} = \dot{x}(x^{\circ(n-1)}(t)) \frac{\delta x^{\circ(n-1)}(t)}{\delta x(\tau)} + \delta(x^{\circ(n-1)}(t)-\tau).
$$
The first term is produced by nothing else than the (functional) chain rule, while the second one is due to the "fixed points" of $x(t)$ together with the rule $\frac{\delta f(x)}{\delta f(y)} = \delta(x-y)$. In the present case, it leads to
$$
\frac{\delta x^{-1}(t)}{\delta x(\tau)} = -\frac{\delta(x^{-1}(t)-\tau)}{\dot{x}(x^{-1}(t))}
$$
when taking $n = 0$, hence finally
$$
\frac{\delta S[x]}{\delta x(\tau)} = \left.\frac{\partial L}{\partial x}\right|_{t=\tau} - \int \left.\frac{\partial L}{\partial x^{-1}}\right|_t \frac{\delta(x^{-1}(t)-\tau)}{\dot{x}(x^{-1}(t))} \,\mathrm{d}t.
$$
In an alternative point of view, it has to be highlighted that the inverse function may be eliminated with the help of a change of variable. Indeed, since $x(t)$ is assumed to be invertible, you might work with the variable $t = f(x)$ instead, hence $x^{-1}(t) = f(t) = f(f(x))$, so that
$$
S[t] = \int L(t,x(t),x^{-1}(t)) \,\mathrm{d}t = \int L(f(x),x,f^2(x)) f'(x)\mathrm{d}x,
$$
where $f^2 = f \circ f$. Consequently, your problem is described equivalently by the following Lagrangian $\tilde{L}(x,f,f^2,f') := f'(x)L(f(x),x,f^2(x))$, where the inverse function has been traded for a second iterate and which can be now extremized with respect to the function $f$.
