This originally comes from $f_1(x,y)=\frac{x}{y}$, where $X=\mathbb{R}^{n}, Y=\mathbb{R}^m, x \in X, f: X \rightarrow Y, x \neq 0, f(x) \neq 0$
$$\limsup_{(h_x,h_y)\to(0,0)} \frac{\left|\frac{x+h_x}{y+h_y}-\frac{x}y\right|} {\sqrt{{h_x}^2+{h_y}^2}}$$
If I understand it correctly, $x$ and $y$ can be anything but zero, and $h_x,h_y$ go towards zero. Moreover, both numerator and denominator cannot be negative.
But since $x,y$ could be anything except for zero, I cannot say whether the numerator or denominator will be greater, so I don't know the result. Is there any way to solve this?
Apply Taylor's formula to the function $f_1$ at the point $(x,y)$. You have $\frac{\partial f_1}{\partial x}(x,y)=\frac{1}y$ and $\frac{\partial f_1}{\partial y}=-\frac{x}{y^2}$. So Taylor's formula gives $$f_1(x+h_x,y+h_y)-f_1(x,y)=\frac{\partial f_1}{\partial x}(x,y)h_x+\frac{\partial f_1}{\partial y}(x,y)h_y+\text{o}(\sqrt{h_x^2+h_y^2}) \\=\frac{1}yh_x-\frac{x}{y^2}h_y+\text{o}(\sqrt{h_x^2+h_y^2}).$$ So now you have $$\frac{\left|f_1(x+h_x,y+h_y)-f_1(x,y)\right|}{\sqrt{h_x^2+h_y^2}} =\left\vert\frac{1}y\frac{h_x}{\sqrt{h_x^2+h_y^2}}-\frac{x}{y^2}\frac{h_x}{\sqrt{h_x^2+h_y^2}}+\text{o}(1)\right\vert.$$ Roughly you have $|a\cos\theta+b\sin\theta|$ where $a=\frac1y$ and $b=-\frac{x}{y^2}$ and you want to try to find $\theta$ which gives you the largest possible value. Can you take it from here?