special relativity - Question about derivation of four-velocity vector

In order to describe a notion of rate of change of positon, in four-dimensional spacetime, we have to introduce the concept of four-velocity.

So, consider the following:

For a massive particle with position $x^{\mu}(t) = (x^{0},x^{1},x^{2},x^{3}) \equiv (x^{0},\vec{x})$ we define the coodinate velocity as:$$ v^{\mu} := \frac{dx^{\mu}}{dt} \equiv (c,\vec{v})\tag{1}$$ Where the spatial components of $(1)$ coincide with classical velocity vector and t is the coordinate time.

But, $(1)$ is not a vector object indeed, because the components didn't transforms as vectors under a lorentz transformation:

$$\frac{dx'^{\mu}}{dt'} = \Lambda^{\mu'}_{\nu}\frac{dx^{\nu}}{dt} \frac{dt}{dt'} = \frac{\Lambda^{\mu'}_{\nu}}{\Lambda^{0'}_{\nu}x^{\nu}}\frac{dx^{\nu}}{dt} \neq \Lambda^{\mu'}_{\nu}\frac{dx^{\nu}}{dt}\tag{2}$$

Well, I simply do not understand one elementary derivation:

--> It's not difficult to know the motivation for this definition:

$$ v^{\mu} := \frac{dx^{\mu}}{dt} \equiv (c,\vec{v})$$

I mean, we need a four-vector and this is certainly a intuitive candidate, but then we realized that this object are not invariant under lorentz transformation, ok. But I'm struggling to derive the expression:

$$\frac{dx^{\mu}}{dt'} = \frac{dx^{\mu}}{dt} \frac{dt}{dt'} \tag{3}$$

Which is important to make the analysis of coordinate transformation as in $(2)$. I know that this is simply the chain rule structure, but I simply do not see how to derive it! The classical chain rule is then:

$$\frac{df[x(t),y(t),z(t)]}{dt} = \frac{\partial f }{\partial x }\frac{dx }{d t}+\frac{\partial f }{\partial y }\frac{d y}{d t}+\frac{\partial f }{\partial y }\frac{d y}{d t}$$

How can I derive $(3)$ from chain rule,explicitly?