To find $\frac{dz}{dt}$, we start with the given function:

$z = \cos(yx^2)$

where $x = t^4 - 2t$ and $y = 1 - t^6$.

First, compute $\frac{dz}{dt}$ using the chain rule:

$\frac{dz}{dt} = -\sin(yx^2) \cdot \frac{d}{dt}(yx^2).$

Next, find the derivative of $yx^2$:

$yx^2 = y \cdot x^2 = (1 - t^6) \cdot (t^4 - 2t)^2.$

Now, use the product rule to differentiate:

$\frac{d}{dt}(yx^2) = \frac{d}{dt}(1 - t^6) \cdot (t^4 - 2t)^2 + (1 - t^6) \cdot \frac{d}{dt}((t^4 - 2t)^2).$

The derivative of $1 - t^6$ is:

$\frac{d}{dt}(1 - t^6) = -6t^5.$

The derivative of $(t^4 - 2t)^2$ is:

Let $u = t^4 - 2t$, then $(t^4 - 2t)^2 = u^2$, and use the chain rule:

$\frac{d}{dt}(u^2) = 2u \cdot \frac{du}{dt}.$

$\frac{du}{dt} = \frac{d}{dt}(t^4 - 2t) = 4t^3 - 2.$

Thus:

$\frac{d}{dt}(u^2) = 2(t^4 - 2t) \cdot (4t^3 - 2).$

Putting it all together, we have:

$\frac{d}{dt}(yx^2) = (-6t^5)(t^4 - 2t)^2 + (1 - t^6)(2(t^4 - 2t)(4t^3 - 2)).$

Then, multiply by $-\sin(yx^2)$ to find $\frac{dz}{dt}$:

$\frac{dz}{dt} = -\sin((1 - t^6)(t^4 - 2t)^2) \left[ (-6t^5)(t^4 - 2t)^2 + (1 - t^6)(2(t^4 - 2t)(4t^3 - 2)) \right].$

This expression is fully in terms of $t$.