The task is to find the partial derivative $f'_y$ (or $\frac{\partial f}{\partial y}$) of the given function at point $M_0 = (2, 0, 4)$. The function is:

$f(x, y, z) = e^{xyz} + (x+1)^3(y+1)^2 + \sin(z-4)$

Step 1: Compute $\frac{\partial f}{\partial y}$

The partial derivative with respect to $y$ is computed term by term:

1. First term: $e^{xyz}$

   • By the chain rule: $\frac{\partial}{\partial y} e^{xyz} = xyz \cdot e^{xyz}$

2. Second term: $(x+1)^3 (y+1)^2$

   • Treat $(x+1)^3$ as a constant: $\frac{\partial}{\partial y} \big((x+1)^3 (y+1)^2\big) = (x+1)^3 \cdot 2(y+1)$

3. Third term: $\sin(z-4)$

   • Since this term does not depend on $y$, its derivative is: $\frac{\partial}{\partial y} \sin(z-4) = 0$

Combining these, the partial derivative is: $\frac{\partial f}{\partial y} = xyz \cdot e^{xyz} + 2(x+1)^3(y+1)$

Step 2: Evaluate at $M_0 = (2, 0, 4)$

Substitute $x = 2$, $y = 0$, and $z = 4$ into $\frac{\partial f}{\partial y}$:

1. First term: $xyz \cdot e^{xyz}$

   • Since $y = 0$, this term becomes: $0 \cdot e^{2 \cdot 0 \cdot 4} = 0$

2. Second term: $2(x+1)^3(y+1)$

   • Substitute $x = 2$ and $y = 0$: $2(2+1)^3(0+1) = 2 \cdot 3^3 \cdot 1 = 2 \cdot 27 = 54$

Adding these, we get: $\frac{\partial f}{\partial y} \Big|_{M_0} = 0 + 54 = 54$

Final Answer:

$f'_y \text{ at } M_0 = (2, 0, 4) \text{ is } \boxed{54}.$

Would you like additional steps explained, or do you have any questions?

Here are some related questions for further practice:

1. What is $\frac{\partial f}{\partial x}$ at $M_0 = (2, 0, 4)$?
2. What is $\frac{\partial f}{\partial z}$ at $M_0 = (2, 0, 4)$?
3. How do you compute higher-order partial derivatives of $f(x, y, z)$?
4. What happens to $f(x, y, z)$ if $x, y, z$ are all equal to 1?
5. Can you explain the physical interpretation of partial derivatives in 3D functions?

Tip: Always double-check if terms vanish due to zero values in derivatives before simplifying!