I have calculated the following Marshallian demand functions
Additionally, I have calculated the indirect utility function for this question
Question: If you take the partial derivative of Pf and divide by the partial derivative of M, the result should be Wf in the first image.Furthermore, if you take the partial derivative of Pc and divide by the partial derivative of M, the result should be Wc in the first image. I cannot seem to get the Wf and Wc when I do this, any assistance to answer this would be greatly appreciate. Cheers!
Given $$U=\sqrt{\frac{M}{P_F \left(\frac{P_F}{P_c}+1\right)}}+\sqrt{\frac{M}{P_c \left(\frac{P_c}{P_F}+1\right)}}$$ To get $W_F$, as you claimed: $$W_F=\frac{\frac{\partial U}{\partial P_F}}{\frac{\partial U}{\partial M}}$$ For $\frac{\partial U}{\partial P_F}, you get:$ $$\begin{align} \frac{\partial U}{\partial P_F}&=\frac{M}{2 P_F^2 \left(\frac{P_c}{P_F}+1\right){}^2 \sqrt{\frac{M}{P_c \left(\frac{P_c}{P_F}+1\right)}}}+\frac{-\frac{M}{P_F^2 \left(\frac{P_F}{P_c}+1\right)}-\frac{M}{P_c P_F \left(\frac{P_F}{P_c}+1\right){}^2}}{2 \sqrt{\frac{M}{P_F \left(\frac{P_F}{P_c}+1\right)}}}\\ &=\frac{1}{2} \left(\frac{M}{\left(P_c+P_F\right){}^2 \sqrt{\frac{M P_F}{P_c \left(P_c+P_F\right)}}}-\frac{\left(P_c+2 P_F\right) \left(\frac{M P_c}{P_F \left(P_c+P_F\right)}\right){}^{3/2}}{M P_c}\right) \end{align}$$ $$\begin{align} \frac{\partial U}{\partial M}&=\frac{1}{2 P_c \left(\frac{P_c}{P_F}+1\right) \sqrt{\frac{M}{P_c \left(\frac{P_c}{P_F}+1\right)}}}+\frac{1}{2 P_F \left(\frac{P_F}{P_c}+1\right) \sqrt{\frac{M}{P_F \left(\frac{P_F}{P_c}+1\right)}}}\\ &=\frac{\sqrt{\frac{M P_F}{P_c \left(P_c+P_F\right)}}+\sqrt{\frac{M P_c}{P_F \left(P_c+P_F\right)}}}{2 M} \end{align}$$ Now as claimed: $$\begin{align} W_F&=\frac{\frac{\partial U}{\partial P_F}}{\frac{\partial U}{\partial M}}\\ &=\frac{\frac{M}{\left(P_c+P_F\right){}^2 \sqrt{\frac{M P_F}{P_c \left(P_c+P_F\right)}}}-\frac{\left(P_c+2 P_F\right) \left(\frac{M P_c}{P_F \left(P_c+P_F\right)}\right){}^{3/2}}{M P_c}}{\frac{2 \left(\sqrt{\frac{M P_F}{P_c \left(P_c+P_F\right)}}+\sqrt{\frac{M P_c}{P_F \left(P_c+P_F\right)}}\right)}{2 M}}\\ &=\frac{M \left(\frac{\sqrt{M} \sqrt{P_c}}{\sqrt{P_F} \left(P_c+P_F\right){}^{3/2}}-\frac{\sqrt{M} \sqrt{P_c} \left(P_c+2 P_F\right)}{P_F^{3/2} \left(P_c+P_F\right){}^{3/2}}\right)}{\frac{\sqrt{M} \sqrt{P_c}}{\sqrt{P_F} \sqrt{P_c+P_F}}+\frac{\sqrt{M} \sqrt{P_F}}{\sqrt{P_c} \sqrt{P_c+P_F}}}\\ &=\color{red}{-}\frac{M P_c}{P_c P_F+P_F^2} \end{align}$$
As claimed: $$W_c=\frac{\frac{\partial U}{\partial P_c}}{\frac{\partial U}{\partial M}}$$ For $\frac{\partial U}{\partial P_c}$, you have: $$\begin{align} \frac{\partial U}{\partial P_c}=&\frac{1}{2} \left(\frac{M}{\left(P_c+P_F\right){}^2 \sqrt{\frac{M P_c}{P_c P_F+P_F^2}}}-\frac{\left(2 P_c+P_F\right) \left(\frac{M P_F}{P_c \left(P_c+P_F\right)}\right){}^{3/2}}{M P_F}\right)\\ \end{align}$$ For $W_c$, you now get: $$\begin{align} W_c=&\frac{\frac{M}{\left(P_c+P_F\right){}^2 \sqrt{\frac{M P_c}{P_c P_F+P_F^2}}}-\frac{\left(2 P_c+P_F\right) \left(\frac{M P_F}{P_c \left(P_c+P_F\right)}\right){}^{3/2}}{M P_F}}{\frac{2 \left(\sqrt{\frac{M P_F}{P_c \left(P_c+P_F\right)}}+\sqrt{\frac{M P_c}{P_F \left(P_c+P_F\right)}}\right)}{2 M}}\\ =&\frac{M \left(\frac{\sqrt{M} \sqrt{P_c P_F+P_F^2}}{\sqrt{P_c} \left(P_c+P_F\right){}^2}-\frac{\sqrt{M} \sqrt{P_F} \left(2 P_c+P_F\right)}{P_c^{3/2} \left(P_c+P_F\right){}^{3/2}}\right)}{\frac{\sqrt{M} \sqrt{P_c}}{\sqrt{P_F} \sqrt{P_c+P_F}}+\frac{\sqrt{M} \sqrt{P_F}}{\sqrt{P_c} \sqrt{P_c+P_F}}}\\ =&\frac{-2 M P_c^2 P_F+P_c \left(M \sqrt{P_F} \sqrt{P_c+P_F} \sqrt{P_F \left(P_c+P_F\right)}-3 M P_F^2\right)-M P_F^3}{P_c \left(P_c+P_F\right){}^3}\\ =&M \left(\frac{\sqrt{P_F} \sqrt{P_F \left(P_c+P_F\right)}}{\left(P_c+P_F\right){}^{5/2}}+\frac{P_c}{\left(P_c+P_F\right){}^2}-\frac{1}{P_c}\right)\\ =&M \left(\frac{P_c}{\left(P_c+P_F\right){}^2}+\frac{P_F}{\left(P_c+P_F\right){}^2}-\frac{1}{P_c}\right)\\-\frac{M P_F}{P_c P_F+P_c^2}\\ =&\color{red}{-}\frac{M P_F}{P_c P_F+P_c^2} \end{align}$$
Though I do not get why your definition of $W_F$ and $W_c$ are without a negative sign.