Statement: If $f_{y}$ exists in a certain neighbourhood of a point (a,b) of the domain of definition of function f, and $f_{yx}$ is continuous at (a,b) then $f_{xy}(a,b)$ exists and is equal to $f_{yx}(a,b)$
Approach:
\begin{align*} f_{xy}=\lim_{h \to 0}\lim_{k \to 0} \frac{\phi(h,k)}{hk}\\ f_{yx}=\lim_{k \to 0}\lim_{h \to 0} \frac{\phi(h,k)}{hk}\\ \end{align*} Where $\phi(h,k) = f(a+h,b+k)-f(a+h,b)-f(a,b+k)+f(a,b) $ and (a,b) is point in the domain of the definition of the function f.
Let $G(x)=f(x,b+k)-f(x,b)$ \begin{align*} \phi(h,k)=G(a+h)-G(a)\\ \text{Applying lagranges mean value theorm: }0 \le\theta \le 1\\ \phi(h,k)=hG^{'}(a+\theta h) \\ =h\{f_{x}(a+\theta h,b+k)-f_{x}(a+\theta h,b)\} \\ \text{Applying lagranges mean value theorm: }0 \le\theta \le 1\\ \phi(h,k)=hkf_{yx}(a+\theta h , b+\theta^{'}k)\\ \Rightarrow \frac{1}{h}\{\frac{f(a+h,b+k)-f(a+h,b)}{k}-\frac{f(a,b+k)-f(a,b)}{k}\} = f_{yx}(a+\theta h , b+\theta^{'}k) \\ \text{taking limit when k tends 0} \\ \frac{f_{y}(a+h,b)-f_{y}(a,b)}{h} = \lim_{k \to 0}f_{yx}(a+\theta h , b+\theta^{'}k) \\ \text{taking limit as h }\to 0\\ f_{xy}(a,b) = \lim_{h \to 0}\lim_{k \to 0}f_{yx}(a+\theta h,b+\theta^{'}k ) =f_{yx}(a,b) \end{align*} I didn't understand this last step. $f_{yx}$ is continuous at (a,b) means $\lim_{(h,k)\to(0,0)}f_{yx}(a+\theta h,b+ \theta^{'} k) = f_{xy}(a,b)$ and if simultaneous limit exist at a point then if the repeated limits exists then these limits are equal but converse is not true. And sometimes even if those repeated limits doesn't exist then also simultaneous limit exists. So how can we show that last step in the proof with out knowing the existence of that repeated limit.