We have an equation (Black-Scholes): $\dfrac{\partial V}{\partial t} + \dfrac{\sigma^2 S^2}{2} \dfrac{\partial^2 V}{\partial S^2} + r S \dfrac{\partial V}{\partial S} - rV = 0$
Let's say we want to differentiate this equation w.r.t. another variable (well, parameter) $\sigma$. What would the result be**? For example
$\dfrac{(\dfrac{\partial V}{\partial t})}{\partial \sigma} \stackrel{?}{=} \dfrac{(\dfrac{\partial V}{\partial \sigma})}{\partial t}$
** Related question, how would the result differ when using a parameter as opposed to a variable?
In general, if $f : \mathbb R^2 \to \mathbb R$ is a continuously differentiable twice, then the order of differentiation of second derivatives is interchangible: $$\partial_{x_1}\partial_{x_2} f = \partial_{x_2}\partial_{x_1}f \quad\text{ where } \partial_{x_i} = \partial/\partial x_i$$
For Black-Scholes, $V$ is assumed to be very well behaved ($C^\infty$) so
$$\partial_t(\partial_\sigma V) = \partial_\sigma(\partial_t V)$$
We can interchange the order of any partial derivatives of $V$ regardless of whether the domain variable it is called a 'parameter' or 'variable'.