Example 1

Consider the function $Z(t) = z(x(t),y(t))$, where
$$z(x,y) = 3x^{\tfrac{1}{3}}y^{\tfrac{2}{3}}, \quad x(t) = 8t^3 \quad \text{and} \quad y(t) = 64t^6.$$
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 $y(t)$:
$$\begin{align} z'_x(x,y) &= 3y^{\tfrac{2}{3}} \cdot \tfrac{1}{3}x^{\tfrac{1}{3}-1} = x^{-\tfrac{2}{3}}y^{\tfrac{2}{3}} = \left(\tfrac{y}{x}\right)^{\tfrac{2}{3}}\\ z'_x(x(t),y(t)) &= \left(\tfrac{64t^6}{8t^3}\right)^{\tfrac{2}{3}} = \left(8t^3\right)^{\tfrac{2}{3}} = 4t^2\\[2mm] z'_y(x,y) &= 3x^{\tfrac{1}{3}} \cdot \tfrac{2}{3}y^{\tfrac{2}{3}-1} = 2x^{\tfrac{1}{3}}y^{-\tfrac{1}{3}} = 2\left(\tfrac{x}{y}\right)^{\tfrac{1}{3}}\\ z'_y(x(t),y(t)) &= 2\left(\tfrac{8t^3}{64t^6}\right)^{\tfrac{1}{3}} = 2\left(\tfrac{1}{8t^3}\right)^{\tfrac{1}{3}} = \tfrac{2}{2t} = \tfrac{1}{t}\\[3mm] x'(t) &= 8 \cdot 3t^2 = 24t^2\\[1mm] y'(t) &= 64 \cdot 6t^5 = 384t^5. \end{align}$$
Now we find $Z'(t)$ by plugging in the prescriptions $z'_x(x(t),y(t))$, $x'(t)$, $z'_y(x(t),y(t))$ and $y'(t)$:
$$Z'(t) = 4t^2 \cdot 24t^2 + \tfrac{1}{t} \cdot 384t^5 = 96t^4 + 384t^4 = 480t^4.$$