Answer
$(4,3)$
Work Step by Step
The slope of the tangent line to the graph of $y=f(x)$ at $(x,f(x))$ is:
$$m_{1}=f'(x)$$
Using the chain rule it follows:
$$m_{1}=(\sqrt{1+2x})'=(1+2x)'\cdot \frac{1}{2\sqrt{1+2x}}=2\cdot \frac{1}{2\sqrt{1+2x}}= \frac{1}{\sqrt{1+2x}}$$
The slope intercept form of the line $6x+2y=1$ is $y=-3x+\frac{1}{2}$.
So the slope of this line is $m_{2}=-3$.
Both lines are perpendicular if the product of their slope is $-1$:
$$\frac{1}{\sqrt{1+2x}} \cdot (-3)=-1$$
$$\frac{1}{\sqrt{1+2x}} \cdot 3=1$$
$$ 3=\sqrt{1+2x}$$
$$ 9=1+2x$$
$$9-1=2x$$
$$8=2x$$
$$4=x$$
So the point is $(4,f(4))$ which is $(4,\sqrt{1+2\cdot 4}=3)$
The point is $(4,3)$.