Answer: Statement

If $z$is a homogeneous function of $x, y$ of order $n,$ then $x {\partial z \over \partial x} +y{\partial z \over \partial y} = nz.$ **Proof.**Since $z$is a homogeneous function of order $n,$ so $z$can be written in the form $z= x^n f(y/x)$.........................(1) Differentiating (1) partially with respect to $'x',$ we have ${\partial z \over \partial x} =nx^{n-1} f(y/x)+x^n f'(y/x)(-y/x^2)$      $=nx^{n-1} f(y/x)-x^{n-2} y f'(y/x).$ Multiplying both sides by $'x',$ we have $x{\partial z \over \partial x} = n x^n f(y/x)-x^{n-1} yf'(y/x)$...............................................................(2) Differentiating (1) partially with respect to $'y',$ we have ${\partial z \over \partial y} = x^n f'(y/x) {1 \over x} = x^{n-1} f'(y/x).$ Multiplying both sides by $y,$ we get  $y{\partial z \over \partial y} = x^{n-1} y f'(y/x).$..............................................................(3) Adding (2) and (3), we have  $x {\partial z \over \partial x} +y{\partial z \over \partial y} = nx^n f(y/x) \Rightarrow x {\partial z \over \partial x} +y{\partial z \over \partial y} =nz.$ Hence proved.