f(x)=y^-(3/2)arctan(x/y).
Consider z=arctan(x/y), then tan(z)=x/y, and sec²(z)∂z/∂x=1/y.
sec²(z)≡1+tan²(z)=1+(x/y)², ∂z/∂x=(1/y)/(1+(x/y)²)=1/(y+x²/y)=y/(x²+y²).
sec²(z)∂z/∂y=-x/y², ∂z/∂y=-(x/y²)/(1+(x/y)²)=-x/(x²+y²).
Consider w=y^-(3/2), ∂w/∂y=-(3/2)y^-(5/2). ∂w/∂x=0.
f(x)=wz,
fᵪ=∂f/∂x=w∂z/∂x+z∂w/∂x=(y^-(3/2))/(y+x²/y) or (y^(-1/2))/(x²+y²).
fᵧ=∂f/∂y=w∂z/∂y+z∂w/∂y=-x(y^-(3/2))/(x²+y²)-((3/2)y^-(5/2))arctan(x/y).