Presumably you mean the second derivative of y with respect to x, d²y/dx².
Compute the first derivative. By the chain rule,
[tex]\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}[/tex]
Differentiate the two parametric equations with respect to t :
[tex]x = \sin(t) \implies \dfrac{dx}{dt} = \cos(t)[/tex]
[tex]y = \cos(t) \implies \dfrac{dy}{dt} = -\sin(t)[/tex]
Then the first derivative is
[tex]\dfrac{dy}{dx} = \dfrac{-\sin(t)}{\cos(t)} = -\tan(t)[/tex]
Now, dy/dx is a function of t, so we can denote it by, say, dy/dx = f(t). Then by the chain rule, the second derivative will be
[tex]\dfrac{d^2y}{dx^2} = \dfrac{df}{dx} = \dfrac{df}{dt} \times \dfrac{dt}{dx} = \dfrac{\frac{df}{dt}}{\frac{dx}{dt}}[/tex]
Differentiating f(t) :
[tex]f(t) = -\tan(t) \implies \dfrac{df}{dt} = -\sec^2(t)[/tex]
Then the second derivative is
[tex]\dfrac{d^2y}{dx^2} = \dfrac{-\sec^2(t)}{\cos(t)} = \boxed{-\sec^3(t)}[/tex]
and since y = cos(t), we can go on to say
[tex]\dfrac{d^2y}{dx^2} = -\dfrac1{y^3}[/tex]
