$$ \mid(x y)=f(x) \cdot f(y) $$ Put $x=y=0$ in (1) to get $f(0)=1$ Put $x=y=1$ in (1) to get $f(1)=0$ or $f(1)=1$ $F(1)=0$ is rejected else $y=1$ in (1) gives $f(x)=0$ Imply $f(0)=0$ Hence, $f(0)=1$ and $f(1)=1$ $$ \begin{aligned} & f^{\prime}(x)=\operatorname{lm}_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\ln _{h \rightarrow 0} f(x)\left(\frac{f\left(1+\frac{h}{x}\right)-f(1)}{h}\right) \\ & =\frac{f(x)}{x} f^{\prime}(1) \\ & =\frac{f^{\prime}(x)}{f(x)}=\frac{k}{x} \Rightarrow \ln \cdot f(x)-k \ln x+c \end{aligned} $$ $$ F(1)=1=\ln 1=k \ln 1+c=c=0 $$ $$ =\ln f(x)=k \ln x \Rightarrow f(x)=x^k \text { but } f(0)=1 $$ $$ \Rightarrow \mathrm{k}=0 $$ $$ \begin{aligned} & \therefore f(x)=1 \\ & \frac{d y}{d x}=f(x)=1 \Rightarrow y=x+c, y(0)=1 \Rightarrow c=1 \\ & \Rightarrow y=x+1 \\ & y\left(\frac{1}{4}\right)+y\left(\frac{3}{4}\right)=\frac{1}{4}+1+\frac{3}{4}+1=3 \end{aligned} $$