Proof
Notice that \begin{align*} W:&=\frac{f(x_0+h,y_0+k)-f(x_0+h,y_0)-f(x_0,y_0+k)+f(x_0,y_0)}{hk}\\ &=\frac{1}{k}\cdot \frac{[f(x_0+h,y_0+k)-f(x_0+h,y_0)]-[f(x_0,y_0+k)-f(x_0,y_0)]}{h}\\ &=\frac{1}{k}\cdot \left[f'_x(x_0+\theta_1 h,y_0+k)-f'_x(x_0+\theta_1h,y_0)\right]\\ &=f''_{xy}(x_0+\theta_1 h,y_0+\theta_2 k). \end{align*} As per the continuity of $f''_{xy}(x,y)$ at $(x_0,y_0)$， $$\lim_{\substack{h \to 0\\ k \to 0}}W=\lim_{\substack{h \to 0\\ k \to 0}}f''_{xy}(x_0+\theta_1 h,y_0+\theta_2 k)=f''_{xy}(x_0,y_0).$$ Likewise，
\begin{align*} W:&=\frac{f(x_0+h,y_0+k)-f(x_0+h,y_0)-f(x_0,y_0+k)+f(x_0,y_0)}{hk}\\ &=\frac{1}{h}\cdot \frac{[f(x_0+h,y_0+k)-f(x_0,y_0+k)]-[f(x_0+h,y_0)-f(x_0,y_0)]}{k}\\ &=\frac{1}{k}\cdot \left[f'_y(x_0+h,y_0+\theta_3 k)-f'_y(x_0,y_0+\theta_3 k)\right]\\ &=f''_{yx}(x_0+\theta_4 h,y_0+\theta_3 k). \end{align*} As per the continuity of $f''_{yx}(x,y)$ at $(x_0,y_0)$， $$\lim_{\substack{h \to 0\\ k \to 0}}W=\lim_{\substack{h \to 0\\ k \to 0}}f''_{yx}(x_0+\theta_4 h,y_0+\theta_3 k)=f''_{yx}(x_0,y_0).$$ It follows that $$f''_{xy}(x_0,y_0)=f''_{yx}(x_0,y_0).$$
It looks correct to me, if a bit technical. One usually obtains this result as a corollary to the fact that $\Delta_y[\Delta_x f](a,b) = \Delta_x[\Delta_y f](a,b) =: \Delta^2_{xy} f(a,b)$, which makes $$\lim_{\Delta x\to 0}\lim_{\Delta y\to 0}\frac{\Delta^2_{xy}f(a,b)}{\Delta x\Delta y}$$ equal to both $f''_{xy}(a,b)$ and $f''_{yx}(a,b)$, avoiding doing your calculations in the numerator twice.