$u =x^y $ Partially differentiating u  w.r.t x, treating y as constant. $\dfrac{\partial u}{\partial x} = y \times x^{y-1}$ Partially differentiating ​ $\dfrac{\partial u}{\partial x}$ w.r.t y, treating x as constant. $\dfrac{\partial}{\partial y} \Big( \dfrac{\partial u}{\partial x}\Big) = x^{y-1} + y x^{y-1} \log x $ (using chain rule, and log here refers to natural logarithm) and since, ​ ​​ $\dfrac{\partial}{\partial y} \Big( \dfrac{\partial u}{\partial x}\Big) =\dfrac{\partial^2 u}{\partial y \ \partial x}$ $\dfrac{\partial^2 u}{\partial y \ \partial x} = x^{y-1} + y x^{y-1} \log x = x^{y-1} (1+y \log x)$ Now, Partially differentiating this w.r.t x, $\dfrac{\partial }{\partial x}\Big(\dfrac{\partial^2 u}{\partial y \ \partial x}\Big) = \dfrac{\partial }{\partial x}( x^{y-1} (1+y \log x)) \ \dfrac{\partial^3 u }{\partial x \ \partial y \ \partial x } \quad = (y-1)x^{y-2} (1+y \log x ) + x^{y-1}\Big(\dfrac{y}{x}\Big) \ \quad \quad \quad \quad = x^{y-2} [(y-1)(1+y \log x ) + y ] \ \quad \quad \quad \quad = x^{y-2}[y-1 +y(y-1)\log x +y] \ \therefore \bbox[5pt]{\dfrac{\partial^3 u }{\partial x \ \partial y \ \partial x } \quad = x^{y-2}\ \Big[2y-1 +y (y-1)\log x \Big]}$