Evaluate all first order partial derivative of $f(x,y,z)=x^\frac{y}{z}$
Here is my attempt:
$y$ and $z$ are constant, so we use the power rule to evaluate $f_x$: $$ f_x = \frac{y}{z}x^{\frac{y-z}{z}}$$
since for any positive constant $c$ the derivative of $f(x)=c^x$ is $f'(x)=c^x \ln c$, we have:
$$ f(x,y,z) = (x^\frac{1}{z})^y \implies f_y = x^\frac{y}{z}\ln x^\frac{1}{z} $$
if we write $g(x)=\frac{1}{x}$ and $h(x)=c^x$ we have that $(h(g(x)))'=g'(x)h'(g(x)) = \frac{-1}{x^2}c^\frac{1}{x}\ln (c)$ by the chain rule. Combining this argument with the previous one we have:
$$f_z = \frac{-1}{z^2}(x^y)^\frac{1}{z}\ln(x^y)$$
of course we can simplify these expressions.
Is my solution right?
For this kind of problems, I think that logarithmic differentiation makes things easy. Let $$f=x^\frac{y}{z}$$ so $$\log(f)=\frac{y}{z}\log(x)$$from which $$\frac{f'_x}{f}=\frac{y}{xz}$$ $$\frac{f'_y}{f}=\frac{\log(x)}{z}$$ $$\frac{f'_z}{f}=-\frac{y}{z^2}\log(x)$$ Continuing using the same kind of approach, you could easily compute higher partial derivatives; for example, $$f'_x=\frac{y}{xz}f$$ $$\log(f'_x)=\log(y)-\log(x)-\log(z)+\log(f)$$ So $$\frac{f''_{xy}}{f'_x}=\frac 1{y}+\frac {f'_y}{f}$$