Incorrect.
We know that:
$\displaystyle \frac{d}{dx}e^x=e^x$
But this time our exponent is $\displaystyle \sec{x}$, not $\displaystyle x$. So what we must do is substitute. Let $\displaystyle u=\sec{x}$. Then:
$\displaystyle e^{\sec{x}}=e^u$
$\displaystyle \frac{d}{du}e^u=e^u=e^{\sec{x}}$
$\displaystyle \frac{dy}{du}=e^{\sec{x}}$
But we can't stop there! What we have taken is $\displaystyle \frac{dy}{du}$. To get $\displaystyle \frac{dy}{dx}$, we have to multiply by $\displaystyle \frac{du}{dx}$, thusly:
$\displaystyle \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$
So, what is $\displaystyle \frac{du}{dx}$? Well, that's simple enough to find:
$\displaystyle u=\sec{x}$
$\displaystyle \frac{du}{dx}=\frac{d}{dx}\sec{x}=\sec{x}\tan{x}$
$\displaystyle \frac{du}{dx}=\sec{x}\tan{x}$
So:
$\displaystyle \frac{dy}{du}\frac{du}{dx}=e^{\sec{x}}\sec{x}\tan{ x}$
And, so, our answer is:
$\displaystyle \frac{dy}{dx}=e^{\sec{x}}\sec{x}\tan{x}$
In the future, you can skip writing down everything and just use this
chain rule in your head.