1. **State the problem:** Find the directional derivative of the function $f(x,y) = x^3 - y^3$ at the point $(4,3)$ in the direction of the vector $\mathbf{v} = \frac{\sqrt{2}}{2}(\mathbf{i} + \mathbf{j})$. 2. **Recall the formula:** The directional derivative of $f$ at point $\mathbf{a} = (x_0,y_0)$ in the direction of a unit vector $\mathbf{u} = (u_1,u_2)$ is given by: $$D_{\mathbf{u}}f(\mathbf{a}) = \nabla f(\mathbf{a}) \cdot \mathbf{u}$$ where $\nabla f$ is the gradient vector of $f$. 3. **Find the gradient $\nabla f(x,y)$:** $$\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = (3x^2, -3y^2)$$ 4. **Evaluate the gradient at $(4,3)$:** $$\nabla f(4,3) = (3 \times 4^2, -3 \times 3^2) = (3 \times 16, -3 \times 9) = (48, -27)$$ 5. **Confirm the direction vector is a unit vector:** Given $\mathbf{v} = \frac{\sqrt{2}}{2}(1,1) = \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right)$. Calculate its magnitude: $$\sqrt{\left( \frac{\sqrt{2}}{2} \right)^2 + \left( \frac{\sqrt{2}}{2} \right)^2} = \sqrt{\frac{2}{4} + \frac{2}{4}} = \sqrt{1} = 1$$ So $\mathbf{v}$ is already a unit vector. 6. **Calculate the directional derivative:** $$D_{\mathbf{v}}f(4,3) = \nabla f(4,3) \cdot \mathbf{v} = (48, -27) \cdot \left( \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) = 48 \times \frac{\sqrt{2}}{2} + (-27) \times \frac{\sqrt{2}}{2}$$ 7. **Simplify:** $$= \frac{\sqrt{2}}{2} (48 - 27) = \frac{\sqrt{2}}{2} \times 21 = 21 \times \frac{\sqrt{2}}{2} = \frac{21\sqrt{2}}{2}$$ **Final answer:** $$\boxed{\frac{21\sqrt{2}}{2}}$$