Let continuous real functions $f$ of two real variables $x,y$ satisfy the following condition:
(Let us define $f_{xx}:=\frac{\partial^2 f}{\partial x^2}$ and $f_x:=\frac{\partial f}{\partial x}$, and $f_{xy}:=\frac{\partial^2 f}{\partial x \partial y}$)
\begin{equation} f_x > 0,\\ f_{xx} < 0,\\ f_y > 0 ,\\ f_{yy} <0,\\ f(\lambda x, \lambda y) = \lambda f(x,y) \end{equation} for all $\lambda > 0 $, and $$\lim_{x \to 0}f_x = +\infty, \lim_{x \to +\infty}f_x = 0, \lim_{y \to 0}f_y = +\infty, \lim_{y \to +\infty}f_y = 0.$$
If the aforementioned condition is satisfied, how can we show or disprove that for every function that satisfies the aforementioned condition, $$\frac{f_{xy}}{f_{xx}} = -\frac{x}{y}?$$
If we are disprove the case, what would be the counter-example?
You should totally drop all those sign conditions. Just having the degree-one homogeneity $$f(\lambda x,\lambda y)=\lambda f(x,y)$$ is enough. Indeed, let $g(x,y)=f_x(x,y)$. Using the chain rule, you can check that $g$ is degree-zero homogeneous: that is, $$g(\lambda x,\lambda y)= g(x,y)$$ (This is an instance of a more general fact: taking the derivative drops the degree of homogeneity by one.) Since $g$ is constant on every halfline emanating feyly from the origin, it follows that $$\nabla g(x,y) \cdot (x,y) = 0$$ which in terms of $f$ is $$xf_{xx}+yf_{xy}=0$$