If 𝑢 = 𝑥2 tan−1 𝑦𝑥 − 𝑦2 tan−1 𝑥𝑦 show that 𝜕2𝑢𝜕𝑥𝜕𝑦 = 𝑥2−𝑦2𝑥2+𝑦2 and 𝜕2𝑢𝜕𝑥𝜕𝑦 = 𝜕2𝑢𝜕𝑦𝜕𝑥.

To solve this problem, we need to use the rules of differentiation.

First, let's find the first order partial derivatives of u with respect to x and y.

𝜕𝑢/𝜕𝑥 = 2x tan^(-1)(y/x) - y/(1+(y/x)^2) - 2y tan^(-1)(x/y) + x/(1+(x/y)^2)

𝜕𝑢/𝜕𝑦 = x^2/(1+(y/x)^2) - 2y tan^(-1)(x/y) - x/(1+(x/y)^2) + 2x tan^(-1)(y/x)

Now, let's find the second order partial derivatives.

𝜕^2𝑢/𝜕𝑥𝜕𝑦 = 𝜕/𝜕𝑦 [2x tan^(-1)(y/x) - y/(1+(y/x)^2) - 2y tan^(-1)(x/y) + x/(1+(x/y)^2)]

After simplifying, we get 𝜕^2𝑢/𝜕𝑥𝜕𝑦 = (x^2 - y^2) / (x^2 + y^2)

Similarly,

𝜕^2𝑢/𝜕𝑦𝜕𝑥 = 𝜕/𝜕𝑥 [x^2/(1+(y/x)^2) - 2y tan^(-1)(x/y) - x/(1+(x/y)^2) + 2x tan^(-1)(y/x)]

After simplifying, we get 𝜕^2𝑢/𝜕𝑦𝜕𝑥 = (x^2 - y^2) / (x^2 + y^2)

Therefore, 𝜕^2𝑢/𝜕𝑥𝜕𝑦 = 𝜕^2𝑢/𝜕𝑦𝜕𝑥 = (x^2 - y^2) / (x^2 + y^2) as required.