Natural deduction proof of $(\alpha\to\beta)\to(\beta\to\gamma)\to(\alpha\to\gamma)$

$$\begin{align} (1) & \alpha \rightarrow \beta && [\text{HYP}] \\ (2) & \beta \rightarrow \gamma && [\text{HYP}] \\ (3) & \alpha && [\text{HYP}] \\ (4) & \beta && [\text{MP}(1,3)] \\ (5) & \gamma && [\text{MP}(2,4)] \\ (6) & \alpha \rightarrow \gamma && [\rightarrow\text{-intro}(3,5)] \\ (7) & (\beta \rightarrow \gamma) \rightarrow (\alpha \rightarrow \gamma) && [\rightarrow\text{-intro}(2,6)] \\ (8) & (\alpha \rightarrow \beta) \rightarrow ((\beta \rightarrow \gamma) \rightarrow (\alpha \rightarrow \gamma)) && [\rightarrow\text{-intro}(1,7)] \\ \end{align}$$

Unless you already have that hypothesis, you need to get to that hypothesis first.

Also, (α→β)→(β→γ)→(α→γ) is ambiguous. You almost surely want to prove ((α→β)→((β→γ)→(α→γ))). Additionally, you don't actually use the deduction (meta) theorem, you use the rule of inference that it implies as valid. I'll call this rule "conditional introduction".

I don't know your format exactly or what you have exactly, but a proof might go something like this:

Hypothesis (rule of modus ponens/detachment): (α→β), α $\vdash$ β

Weakening: (α→β), α, (β→γ) $\vdash$ β

Commutation on the left side: (α→β), (β→γ), α $\vdash$ β

Identity: (α→β), (β→γ), α $\vdash$ β, (β→γ)

Detachment: (α→β), (β→γ), α $\vdash$ γ

Conditional introduction: (α→β), (β→γ) $\vdash$ (α→γ)