Find an equation for the straight line tangent to this curve at the given point. As you can check, the point $(1,\ln(2))$ lies on the curve
$xe^y+7y−2x=7\ln(2).$
Find an equation for the straight line tangent to this curve at the given point.
Your answer should be an equation involving x and/or y.
Use ln() for the natural logarithm.
Tried to fin a function y=... but not sure how to do it.
 A: You have a point on the curve: $(x_0, y_0) = (1, \ln 2)$. You need to determine the "slope" $m$ of the tangent line at $x = 1$. To find that slope, you need to find derivative by implicit differentiation: $$\dfrac{2 - e^y}{xe^y + 7}$$ then evaluate at $x = 1, y = \ln 2$, and set $f'(1, \ln 2) = m = 0$.
Then you'll have both a point on the line, and slope, and using point-slope form for an equation, you're good to go:
$$y - y_0 = m(x - x_0)$$
$$y - \ln 2 = 0(x - 1) = 0 \iff y = \ln 2$$ And you've got your line, parallel to the x-axis.
A: The curve, if I read it right, is
$xe^y + 7y -2x = 7 \ln 2, \tag{1}$
and sure enough, taking $(x, y) = (1, \ln 2)$ provides a point on this curve.
We differentiate (1) implicitly, yielding
$e^y + xy'e^y + 7y' - 2 = 0, \tag{2}$
and solve the resulting equation for $y'$; it's linear in $y'$, so that helps:
$e^y + y'(xe^y + 7) = 2, \tag{3}$
so
$y' = (2 - e^y)(xe^y + 7)^{-1}. \tag{4}$
Plugging in $(x, y) = (1, \ln 2)$ gives
$y' = (2 - 2)(2 + 7) = 0 \tag{5}$
at the point $(1, \ln 2)$.
So using the general form $(y - y_0) = m(x - x_0)$ for the tangent line, as suggested by AmWhy in his answer, gives
$y - \ln 2 = 0; \tag{5}$
so the tangent line is horizontal at the point $(1, \ln 2)$!
Hope this helps!  Cheers, and as always, 
Fiat Lux
