Consider the function $Z(t) = z(x(t),y(t))$, where
$$z(x,y) = 2xy^2, \quad x(t) = \ln(t) \quad \text{and} \quad y(t) = 3t^2.$$
Determine $Z'(t)$.
According to the chain rule (case 1)
$$Z'(t) = z'_x(x(t),y(t))x'(t) + z'_y(x(t),y(t))y'(t).$$
Hence, we need the partial derivatives of $z(x,y)$ at $(x(t),y(t))$, the derivative of $x(t)$ and the derivative of $y(t)$:
$$
\begin{align}
z'_x(x,y) &= 2y^2 \cdot 1 = 2y^2\\
z'_x(x(t),y(t)) &= 2\left(3t^2\right)^2 = 2\cdot 9t^4 = 18t^4\\[2mm]
z'_y(x,y) &= 2x \cdot 2y = 4xy\\
z'_y(x(t),y(t)) &= 4 \cdot \ln(t) \cdot 3t^2 = 12t^2\ln(t)\\[2mm]
x'(t) &= \tfrac{1}{t}\\[1mm]
y'(t) &= 3 \cdot 2t = 6t.
\end{align}
$$
Now we find $Z'(t)$ by plugging in the prescriptions of $z'_x(x(t),y(t))$, $x'(t)$, $z'_y(x(t),y(t))$ and $y'(t)$:
$$
Z'(t) = 18t^4 \cdot \tfrac{1}{t} + 12t^2\ln(t) \cdot 6t = 18t^3 + 72t^3\ln(t).
$$
