Take the function $y(\vec v)$ such that $y:\mathbb R^n\to\mathbb R$. Given it's gradient $\nabla y = \left(\frac{\partial y}{\partial v_1},\cdots,\frac{\partial y}{\partial v_n}\right)$, it is possible to state that for a real constant $\lambda$,
$$ \nabla \left(\frac 1\lambda\cdot y\right) = \frac 1\lambda\cdot\nabla y $$
But supposing that the function $y(\vec v)$ can be represented by a derivative $\frac{dz}{dx}$, where x is a scalar. Then the gradient becomes $\nabla \frac{dz}{dx}$. Would it be correct to state the following, or can I not treat infinitesimals like real numbers?
$$\nabla \frac{dz}{dx} = \frac{1}{dx}\cdot\nabla dz$$
Much appreciated.
In the background there is a scalar function $$z:\>{\mathbb R}^n\times{\mathbb R}\to{\mathbb R},\qquad(v,t)\mapsto z(v,t)$$ of $n$ space variables $v_k$ and time $t$. Suppose that $$z(\cdot,\cdot)\in C^2({\mathbb R}^{n+1})\tag{1}$$ to simplify matters. The time derivative $$y(v,t):={\partial z\over\partial t}(v,t)$$ of $z(\cdot,\cdot)$ is again a function of the space variables $v_k$ and $t$. At the beginning of your post you are talking about the gradient of $y(\cdot,\cdot)$, which is by definition the (time dependent) vector field $$\nabla y(v,t):=\left({\partial y(v,t)\over\partial v_1},\ldots, {\partial y(v,t)\over\partial v_n}\right)$$ (the $\nabla$ operator only acts on the space variables).
The question now is whether this gradient $\nabla y(v,t)$ can be considered as the time derivative of the gradient $\nabla z(v,t)$. The answer is Yes; one has $$\nabla y(v,t)={\partial\over\partial t}\nabla z(v,t)\qquad\bigl((v,t)\in{\mathbb R}^n\times{\mathbb R}\bigr) .$$ The reason is simple: The assumption $(1)$ guarantees that "mixed partials" are equal. Therefore it doesn't matter whether we first differentiate with respect to a $v_k$ or with respect to $t$.