u = log( $x^2$ + $y^2$ )

$ \therefore \frac{\partial u}{\partial x} = \frac{\partial \; }{\partial x} \Big[ log( x^2 \; \; + \; \; y^2 ) \Big] \; = \; \frac{2x}{x^2 \; \; + \; \; y^2} \; \; \; \ldots \; (i) \\ Similarly, \; \frac{\partial u}{\partial y} = \frac{\partial \; }{\partial y} \Big[ log( x^2 \; \; + \; \; y^2 ) \Big] \; = \; \frac{2y}{x^2 \; \; + \; \; y^2} \\ Now, \; \frac{\partial^2 \; u}{\partial x \partial y} \; = \; \frac{\partial }{\partial x} \bigg( \frac{\partial u }{\partial y} \bigg) \; = \; \frac{\partial }{\partial x} \bigg[ \frac{2y}{x^2 \; \; + \; \; y^2} \bigg] \; \; \; \; \ldots From \; \; (i) \\ = \frac{ (x^2 \; \; + \; \; y^2) \frac{\partial }{\partial x} (2y) \; \; - \; \; 2y \frac{\partial }{\partial x} (x^2 \; \; + \; \; y^2) }{ (x^2 \; \; + \; \; y^2)^2 } \\ \; \\ \therefore \frac{\partial^2 \; u}{\partial x \partial y} \; = \; \frac{ 0 \; - \; 2y(2x)}{ (x^2 \; \; + \; \; y^2)^2 } \; = \; \frac{ - 4xy }{ (x^2 \; \; + \; \; y^2)^2 } \; \; \; \; \; \; \ldots (ii) \\ \; \\ Now, \; \frac{\partial^2 \; u}{\partial y \partial x} \; = \; \frac{\partial }{\partial y} \bigg( \frac{\partial u }{\partial x} \bigg) \; = \; \frac{\partial }{\partial y} \bigg[ \frac{2x}{x^2 \; \; + \; \; y^2} \bigg] \; \; \; \; \ldots From \; \; (i) \\ = \frac{ (x^2 \; \; + \; \; y^2) \frac{\partial }{\partial y} (2x) \; \; - \; \; 2x \frac{\partial }{\partial y} (x^2 \; \; + \; \; y^2) }{ (x^2 \; \; + \; \; y^2)^2 } \\ \; \\ = \frac{ 0 \; - \; 2x(2y)}{ (x^2 \; \; + \; \; y^2)^2 } \\ \therefore \frac{\partial^2 \; u}{\partial y \partial x} \; = \; \frac{ - 4xy }{ (x^2 \; \; + \; \; y^2)^2 } \; \; \; \; \ldots (iii) \\ \; \\ From \; (ii) \; and \; (iii), \; \; it \; is \; proved \; \; that \; \; \frac{\partial^2 \; u}{\partial x \partial y} \; = \; \frac{\partial^2 \; u}{\partial y \partial x} $