This is only true if ##f## is a function of two variables and there is a function ##g## of one variable such that ##y=g(x)##.
Then we have ##f(x,y)=f(x,g(x))## so then, using the definitions of derivatives as limits, we can write:
\begin{align*}
\frac d{dx}f(x,y)&=\frac d{dx}f(x,g(x))\\
&=\lim_{h\to 0}\left[\frac{f(x+h,g(x+h))-f(x,g(x))}{h}\right]\\
&=\lim_{h\to 0}\left[
\frac{f(x+h,g(x+h))-f(x,g(x+h))}{h}
+\frac{f(x,g(x+h))-f(x,g(x))}{h}
\right]\\
&=\lim_{h\to 0}\left[
\frac{f(x+h,g(x+h))-f(x,g(x+h))}{h}
\right]+
\lim_{h\to 0}\left[
\frac{f(x,g(x+h))-f(x,g(x))}{h}
\right]
\end{align*}
provided both limits exist.
The first term is a bit tricky. Provided certain assumptions hold, the details of which we won't go into here, it is equal to the double limit:
\begin{align*}
\lim_{h'\to 0}\bigg[
\lim_{h\to 0}\bigg(
&\frac{f(x+h,g(x+h'))-f(x,g(x+h'))}{h}
\bigg)
\bigg]\\
&=\lim_{h'\to 0}\bigg[
\frac{\partial}{\partial x}f(x,g(x+h'))
\bigg]\\
&=
\frac{\partial}{\partial x}f(x,g(x))\\
\end{align*}
The second term is equal to:
\begin{align*}
\lim_{h\to 0}\bigg[
\frac{f(x,g(x+h))-f(x,g(x))}{g(x+h)-g(x)}
&\times
\frac{g(x+h)g(x)}{h}
\bigg]\\
&=\lim_{h\to 0}\left[
\frac{f(x,g(x+h))-f(x,g(x))}{g(x+h)-g(x)}\right]
\times
\lim_{h\to 0}\left[\frac{g(x+h)-g(x)}{h}
\right]\\
&=\lim_{\delta y\to 0}\left[
\frac{f(x,y+\delta y)-f(x,y)}{\delta y}\right]
\times
\lim_{h\to 0}\left[\frac{g(x+h)-g(x)}{h}
\right]\\
&=
\frac{\partial }{\partial y}f(x,y)
\times
\frac{dg}{dx}(x)\\
&=
\frac{\partial }{\partial y}f(x,y)
\times
\frac{dy}{dx}\\
\end{align*}
Summing the terms, we get
$$\frac{d}{dx}f(x,y)=\frac{\partial}{\partial x}f(x,y)+\frac{\partial }{\partial y}f(x,y)
\times
\frac{dy}{dx}$$
as required.
