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.