Let 𝑢 = 𝑓(𝑥, 𝑦) = (𝑒^𝑥)𝑠𝑒𝑛(3𝑦). Check if 9((𝜕^2) u / 𝜕(𝑥^2)) +((𝜕^2) 𝑢 / 𝜕(𝑦^2)) = 0

To check if $$\(9\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0\)$$, we need to calculate the second partial derivatives of u with respect to x and y, and then verify if the equation holds. First, let's find the first and second partial derivatives of $$\(u = f(x, y) = e^x \sin(3y)\)$$: 1. First partial derivatives: $$\(\frac{\partial u}{\partial x} = \frac{\partial}{\partial x}(e^x \sin(3y))\)$$ Using the product rule, we have: $$\(\frac{\partial}{\partial x}(e^x \sin(3y)) = e^x \sin(3y) + e^x \frac{\partial}{\partial x}(\sin(3y))\)$$ $$\(\frac{\partial u}{\partial x} = e^x \sin(3y) + 0\)$$ $$ \(\frac{\partial u}{\partial x} = e^x \sin(3y)\)$$ 2. Now, let's find the second partial derivative with respect to x: $$\(\frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x}(e^x \sin(3y))\)$$ Using the product rule again: $$\(\frac{\partial}{\partial x}(e^x \sin(3y)) = e^x \sin(3y) + e^x \frac{\partial}{\partial x}(\sin(3y))\)$$ The second term, $$\(\frac{\partial}{\partial x}(\sin(3y))\)$$, is 0 since the derivative of sin(3y) with respect to x is 0. So, $$\(\frac{\partial^2 u}{\partial x^2} = e^x \sin(3y)\)$$ 3. Next, let's find the first partial derivative with respect to y: $$\(\frac{\partial u}{\partial y} = \frac{\partial}{\partial y}(e^x \sin(3y))\)$$ Using the chain rule, we have: $$\(\frac{\partial}{\partial y}(e^x \sin(3y)) = e^x \cdot 3\cos(3y)\)$$ $$\(\frac{\partial u}{\partial y} = 3e^x\cos(3y)\)$$ 4. Finally, let's find the second partial derivative with respect to y: $$\(\frac{\partial^2 u}{\partial y^2} = \frac{\partial}{\partial y}(3e^x\cos(3y))\)$$ Using the chain rule again: $$\(\frac{\partial}{\partial y}(3e^x\cos(3y)) = 3e^x\cdot(-3\sin(3y)) = -9e^x\sin(3y)\)$$ Now, we can plug these results into the equation $$\(9\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0\)$$: $$\(9(e^x \sin(3y)) + (-9e^x\sin(3y)) = 0\)$$ $$\(9e^x \sin(3y) - 9e^x\sin(3y) = 0\)$$ The equation simplifies to: $$\(0 = 0\)$$ So, the equation $$\(9\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0\)$$ holds, and it is satisfied.