By definition of homogeneity, if $f(x, y)$ is homogeneous of degree $n$, then $$f(tx, ty) = t^nf(x, y)$$
Problem
Prove that if $f(x, y)$ is homogeneous of degree $1$, then $f_{xx}(x, y) f_{yy}(x, y) = (f_{xy})^2$
Attempted solution
I tried using Euler's theorem on homogeneous functions, which asserts that if $f(x, y)$ is homogeneous of degree $n$, then $$xf_x(x, y) + yf_y(x, y) = nf(x, y)$$
But this didn't lead me to the desired result:
$$xf_x(x, y) + yf_y(x, y) = nf(x, y)$$ $$\frac{xf_x(x, y)}{nf(x, y)} + \frac{yf_y(x, y)}{nf(x, y)} = 0$$ $$???$$
By the title, you have that $f(x,y)$ is homogeneus of degree 1, so in particular you have $$xf_{x}(x,y)+yf_{y}(x,y)=f(x,y)$$ You can try deriving that expression by $x$ and by $y$ so you get the form of $f_{xx}$ and $f_{yy}$