Proving 2nd ode

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If $x=e^t$, can someone give me proof that $$ \frac{d^2}{dx^2}=\frac{1}{e^{2t}}\left(\frac{d^2}{dt^2}−\frac{d}{dt}\right). $$ Thank you

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$$ \frac{d}{dx} = \frac{d}{dt} \times \frac{dt}{dx} = \frac{d/dt}{dx/dt} = e^{-t} \frac{d}{dt}. $$

Note that $$ \frac{d^2}{dx^2} = \frac{d}{dx} \left[ \frac{d}{dx} \right] $$ and you can now use the previous result.