If {z^2} = frac{{x^{1/2} + y^{1/2}}}{{x^{1/3} | vedclass

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(D) Let $f(x, y) = z^2 = \frac{x^{1/2} + y^{1/2}}{x^{1/3} + y^{1/3}}$.Check for homogeneity: $f(tx, ty) = \frac{(tx)^{1/2} + (ty)^{1/2}}{(tx)^{1/3} +

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$\frac{z}{12}$

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(D) Let $f(x, y) = z^2 = \frac{x^{1/2} + y^{1/2}}{x^{1/3} + y^{1/3}}$.Check for homogeneity: $f(tx, ty) = \frac{(tx)^{1/2} + (ty)^{1/2}}{(tx)^{1/3} + (ty)^{1/3}} = \frac{t^{1/2}(x^{1/2} + y^{1/2})}{t^{1/3}(x^{1/3} + y^{1/3})} = t^{1/2 - 1/3} f(x, y) = t^{1/6} f(x, y)$.Since $z^2$ is a homogeneous function of degree $n = \frac{1}{6}$,by Euler's Theorem on homogeneous functions:$x \frac{\partial (z^2)}{\partial x} + y \frac{\partial (z^2)}{\partial y} = n z^2 = \frac{1}{6} z^2$.Using the chain rule,$\frac{\partial (z^2)}{\partial x} = 2z \frac{\partial z}{\partial x}$ and $\frac{\partial (z^2)}{\partial y} = 2z \frac{\partial z}{\partial y}$.Substituting these into the equation:$x(2z \frac{\partial z}{\partial x}) + y(2z \frac{\partial z}{\partial y}) = \frac{1}{6} z^2$.Dividing both sides by $2z$ (assuming $z 
eq 0$):$x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = \frac{1}{12} z$.

If ${z^2} = \frac{{x^{1/2} + y^{1/2}}}{{x^{1/3} + y^{1/3}}}$,then $x\frac{{\partial z}}{{\partial x}} + y\frac{{\partial z}}{{\partial y}} = $

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