Unit vector for the minimum directional derivative of a function

For the following function: $$f(x,y)=\ln(x-y)$$, at $$(2,1)$$ what is the unit vector for the minimum directional derivative of $$f$$ ?

Firstly to find the gradient of $$f$$ at $$(2,1)$$: $$\nabla f(x,y) = \begin{bmatrix} \frac{\partial}{\partial x}(\ln(x-y))\\ \frac{\partial}{\partial y}(\ln(x-y))\\ \end{bmatrix} = \begin{bmatrix} \frac{1}{x-y}\\ \frac{-1}{x-y}\\ \end{bmatrix}$$

$$\nabla f(2,1) = \begin{bmatrix} 1\\ -1\\ \end{bmatrix} = \langle 1,-1\rangle$$

The minimum directional derivative is: $$-|\nabla f(2,1)| = -|\langle 1,-1\rangle|=-\sqrt 2$$

To find the unit vector: $$u = \frac{\nabla f(2,1)}{-|\nabla f(2,1)|} = \frac{\langle 1,-1\rangle}{-\sqrt2} = \frac{\langle -1,1\rangle}{\sqrt2}$$

$$\therefore$$ The minimum directional derivative of $$f(x,y)$$ is $$-\sqrt 2$$ and occurs in the direction of unit vector $$\frac{\langle -1,1\rangle}{\sqrt2}$$

Is this the correct procedure to solve the problem? Thanks in advance.

As an example : it seems to me that you're implicitly using the property that the directional derivative $$\partial_v f$$ at a point $$a$$ is the dot product $$\nabla f (a) \cdot v$$. This is not always the case ! But it is true when the function $$f$$ is differentiable at $$a.$$ So your proof should at least contain the words " Since $$f$$ is differentiable at ..."
The minimum directional derivative is: $$−|\nabla f(2,1)|=−|⟨1,−1⟩|=−\sqrt 2$$