Let's solve each of the partial derivatives step-by-step for the given function:

Given: $f(x, y) = -3x^6 + 5x^2 y^3 + 3y^2$

We are asked to find the following derivatives:

Step 1: Calculate $f_x(x, y)$ (Partial derivative with respect to $x$)

Differentiate $f(x, y)$ with respect to $x$, treating $y$ as a constant.

$f_x(x, y) = \frac{\partial}{\partial x}(-3x^6 + 5x^2 y^3 + 3y^2)$

Calculating each term:

• For $-3x^6$: $\frac{\partial}{\partial x}(-3x^6) = -18x^5$
• For $5x^2 y^3$: $\frac{\partial}{\partial x}(5x^2 y^3) = 10x y^3$
• For $3y^2$: $\frac{\partial}{\partial x}(3y^2) = 0$ (since it is independent of $x$)

So, $f_x(x, y) = -18x^5 + 10x y^3$

Step 2: Calculate $f_y(x, y)$ (Partial derivative with respect to $y$)

Differentiate $f(x, y)$ with respect to $y$, treating $x$ as a constant.

$f_y(x, y) = \frac{\partial}{\partial y}(-3x^6 + 5x^2 y^3 + 3y^2)$

Calculating each term:

• For $-3x^6$: $\frac{\partial}{\partial y}(-3x^6) = 0$ (since it is independent of $y$)
• For $5x^2 y^3$: $\frac{\partial}{\partial y}(5x^2 y^3) = 15x^2 y^2$
• For $3y^2$: $\frac{\partial}{\partial y}(3y^2) = 6y$

So, $f_y(x, y) = 15x^2 y^2 + 6y$

Step 3: Calculate $f_{xx}(x, y)$ (Second partial derivative with respect to $x$)

Differentiate $f_x(x, y)$ with respect to $x$ again.

$f_{xx}(x, y) = \frac{\partial}{\partial x}(-18x^5 + 10x y^3)$

Calculating each term:

• For $-18x^5$: $\frac{\partial}{\partial x}(-18x^5) = -90x^4$
• For $10x y^3$: $\frac{\partial}{\partial x}(10x y^3) = 10 y^3$

So, $f_{xx}(x, y) = -90x^4 + 10 y^3$

Step 4: Calculate $f_{xy}(x, y)$ (Mixed partial derivative with respect to $x$ and $y$)

Differentiate $f_x(x, y)$ with respect to $y$.

$f_{xy}(x, y) = \frac{\partial}{\partial y}(-18x^5 + 10x y^3)$

Calculating each term:

• For $-18x^5$: $\frac{\partial}{\partial y}(-18x^5) = 0$ (independent of $y$)
• For $10x y^3$: $\frac{\partial}{\partial y}(10x y^3) = 30x y^2$

So, $f_{xy}(x, y) = 30x y^2$

Summary of Results

• $f_x(x, y) = -18x^5 + 10x y^3$
• $f_y(x, y) = 15x^2 y^2 + 6y$
• $f_{xx}(x, y) = -90x^4 + 10 y^3$
• $f_{xy}(x, y) = 30x y^2$

Would you like any further details or explanations on these steps?

Here are five follow-up questions related to this problem:

1. How would you find $f_{yy}(x, y)$, the second partial derivative with respect to $y$?
2. What is the significance of the mixed partial derivative $f_{xy}$ in multivariable calculus?
3. How does the function $f(x, y)$ behave for large values of $x$ and $y$?
4. Can you interpret the meaning of these partial derivatives in a physical context?
5. What is the importance of second partial derivatives in determining concavity and convexity?

Tip: In multivariable calculus, when calculating mixed partial derivatives, be consistent with differentiating terms involving both variables step-by-step to avoid errors.