2. Find the domain of the following functions: a) \( f(x, y)=\frac{2 x-3 y}{\sqrt[3]{4 x^{2}+9 y^{2}}} \) b) \( g(x, y)=\frac{1}{x^{2}-6 x y+9 y^{2}} \) c) \( h(x, y)=\log \left(\frac{x^{2}}{x-y}\right) \)

**(a)** We have \[ f(x,y)=\frac{2x-3y}{\sqrt[3]{4x^2+9y^2}}. \] The cube root function is defined for all real numbers. However, since the cube root appears in the denominator, we must have \[ \sqrt[3]{4x^2+9y^2}\neq 0. \] Since \(\sqrt[3]{4x^2+9y^2}=0\) if and only if \[ 4x^2+9y^2=0, \] and because \(4x^2\) and \(9y^2\) are nonnegative, the equality holds only when \[ x=0 \quad \text{and} \quad y=0. \] Thus, the domain is all \((x,y)\in\mathbb{R}^2\) except \((0,0)\). --- **(b)** We consider \[ g(x,y)=\frac{1}{x^2-6xy+9y^2}. \] Notice that the quadratic in the denominator can be factored as \[ x^2-6xy+9y^2=(x-3y)^2. \] Since division by zero is not allowed, we require \[ (x-3y)^2\neq 0, \] which is equivalent to \[ x-3y\neq 0. \] Thus, the domain is all \((x,y)\in\mathbb{R}^2\) such that \[ x\neq 3y. \] --- **(c)** We are given \[ h(x,y)=\log\left(\frac{x^2}{x-y}\right). \] For the logarithm to be defined, its argument must be positive: \[ \frac{x^2}{x-y}>0. \] Since \(x^2\ge 0\) for all \(x\) and equals zero if and only if \(x=0\), in order for the quotient to be positive, we must have: 1. \(x^2>0\) (i.e. \(x\neq 0\)), and 2. The sign of \(x-y\) must be positive (because if \(x^2>0\) then the sign of the quotient is the sign of \(x-y\)). Thus, we require \[ x-y>0\quad \text{and} \quad x\neq 0. \] Notice that the inequality \(x-y>0\) is equivalent to \(x>y\). --- **Summary of Answers:** - **(a)** The domain is \[ \{(x,y)\in\mathbb{R}^2 : (x,y)\neq (0,0)\}. \] - **(b)** The domain is \[ \{(x,y)\in\mathbb{R}^2 : x\neq 3y\}. \] - **(c)** The domain is \[ \{(x,y)\in\mathbb{R}^2 : x\neq 0 \text{ and } x>y\}. \]