Is my proof correct?
Let a and b defined such that: $$\sqrt{a^2+b^2} = 1 \\<br /> \frac {df}{d(x, y)} = \lim_{h \to 0} \frac{f(x+ah, y+bh) - f(x,y)}{h\sqrt{a^2+b^2}} = \lim_{h \to 0} \frac{f(x+ah, y+bh) - f(x,y)}{h} \\<br /> \frac {df}{dx} = \lim_{h \to 0} \frac{f(x+h, y) - f(x,y)}{h} \\<br /> \frac {d_af}{dx} = \lim_{h \to 0} \frac{f(x+ah, y) - f(x,y)}{h}$$
$$\mathbf{Theorem \; 1 \;} a\frac{df}{dx} + b\frac{df}{dy} = \frac{df}{d(x,y)} \\<br /> \mathbf{Proof:}$$
$$\mathbf{Theorem \; 1.1 \;} \frac {d_af}{dx} = a\frac{df}{dx} \\<br /> \mathbf{Proof:}$$
Let $g(z) = az$ where $z = \frac{x}{a}$
Let $h(z) = f(g(z), y)$ (such that $h'(z) = \frac{d_af}{dx}$)
$$h'(z) = \lim_{\epsilon \to 0} \frac{h(z+\epsilon) - h(z)} {\epsilon} \\<br /> h'(z) = \lim_{\epsilon \to 0} \frac{f(g(\frac{x}{a}+\epsilon),y) - f(g(\frac{x}{a}), y)}{\epsilon} \\<br /> h'(z) = \lim_{\epsilon \to 0} \frac{f(x+ah, y) - f(x, y))}{\epsilon} \\<br /> h'(z) = \frac{d_af}{dx} (as \; desired)$$
Or equivalently using the chain rule:
$$h'(z) = f'(g(z),y) f(g'(z), y) \\<br /> g'(z) = a \\<br /> \therefore h'(z) = a f'(g(z),y) \\<br /> h'(z) = a \lim_{\epsilon \to 0} \frac{f(g(\frac x a)+\epsilon, y) - f(g(x/a), y))}{\epsilon} \\<br /> h'(z) = a \lim_{\epsilon \to 0} \frac{f(x+\epsilon, y) - f(x,y)}{\epsilon} \\<br /> \therefore \frac {d_af}{dx} = a\frac{df}{dx} \square$$
$$\mathbf{Theorem \; 1.2 \;} \frac{df}{d(x,y)} = \frac{d_af}{dx} + \frac{d_bf}{dy} \\<br /> \mathbf{Proof:} \\<br /> a\frac{df}{dx} = a\frac{df}{dx} \\<br /> \lim_{h \to 0} \frac{f(x+ah, y+bh) - f(x,y+bh)}{h} = \lim_{h \to 0} \frac{f(x+ah,y) - f(x,y)}{h} + \frac{f(x,y) - f(x,y)}{h} (by \; Theorem \; 1) \\<br /> \lim_{h \to 0} \frac{f(x+ah,y+bh) - f(x,y)}{h} = \lim_{h \to 0} \frac{f(x + ah, y) - f(x,y)}{h} + \frac{f(x,y+bh) - f(x,y)}{h} \\<br /> \therefore \frac{df}{d(x,y)} = \frac{d_af}{dx} + \frac{d_bf}{dy} \square \\<br /> \therefore \frac{df}{d(x,y)} = a\frac{df}{dx} + b\frac{df}{dy} (by \; Theorem \; 1.1 \; and \; 1.2) \blacksquare$$