Answer
$y'=-\dfrac{3(\sec^{2}3x)(e^{\sqrt{\tan3x}})\sin(e^{\sqrt{\tan3x}})}{2\sqrt{\tan3x}}$
Work Step by Step
$y=\cos(e^{\sqrt{\tan3x}})$
Start the differentiation process by using the chain rule:
$y'=-\sin(e^{\sqrt{\tan3x}})(e^{\sqrt{\tan3x}})'=...$
Apply the chain rule one more time to evaluate the indicated derivative:
$...=-(e^{\sqrt{\tan3x}})\sin(e^{\sqrt{\tan3x}})(\sqrt{\tan3x})'=...$
$...=-(e^{\sqrt{\tan3x}})\sin(e^{\sqrt{\tan3x}})[(\tan3x)^{1/2}]'=...$
Once again, apply the chain rule to evaluate the indicated derivative:
$...=-\dfrac{1}{2}(\tan3x)^{-1/2}(\tan3x)'(e^{\sqrt{\tan3x}})\sin(e^{\sqrt{\tan3x}})=...$
Apply the chain rule one last time to evaluate the indicated derivative and simplify the expression:
$...=-\dfrac{1}{2}(3x)'(\tan3x)^{-1/2}(\sec^{2}3x)(e^{\sqrt{\tan3x}})\sin(e^{\sqrt{\tan3x}})=...$
$...=-\dfrac{3(\sec^{2}3x)(e^{\sqrt{\tan3x}})\sin(e^{\sqrt{\tan3x}})}{2\sqrt{\tan3x}}$
