Find an equation of the tangent line to the following graph at the given pont: $ \ln(x+1)+e^{x+y^2}=1$, at$ (0,0)$ I tried to differentiate it then use the equation of the tangent line once I get the slope after differentiation, but I was told my answer was wrong.
 A: Let $x=x(y)$, then after differentiating we get
$$\frac{x'(y)}{x(y)+1}+e^{x+y^2}\cdot (x'(y)+2y)=0.$$
For $y=0$ we obtain $2x'(0)=0$. Hence the tangent line at $(0,0)$ is $x=0$.
P.S. Note that from $\ln(x+1)+e^{x+y^2}=1$, it is easy to see that
$$y(x)=\pm \sqrt{\ln\left(\frac{1-\ln(x+1)}{e^x} \right)}$$
which means that the curve can not be considered as the graph of a function $y(x)$ in a neighborhood of $(0,0)$. 
A: Okay, say the point we are given is $(x_1,y_1)$, where $\displaystyle \ln(x_1+1)+e^{x_1+y_1^2}=1$
Using implicit derivation, 
$$\frac{1}{x+1}+e^{x+y^2}\cdot(1+2yy')=0$$
$$e^{x+y^2}\cdot(1+2yy')=-\frac{1}{x+1}$$
$$(1+2yy')=-\frac{e^{-x-y^2}}{(x+1)}$$
$$y'=-\frac{\frac{e^{-x-y^2}}{(x+1)}+1}{2y}$$
So the line you are looking for is 
$$y-y_1=-\frac{\frac{e^{-x-y^2}}{(x+1)}+1}{2y}\cdot(x-x_1)$$
Plug in $y_1=x_1=0$. You will obtain $x=0$.
A: First note that $ ln(x+1)+e^{x+y^2}=1$ is the equation of a line but it is not an implicit function $y=f(x)$. Rewrite the equation as:
$$
\ln(x+1)+e^xe^{y^2}=1
$$
and solve for $y$:
$$
y^2=\ln\left[\frac{1-\ln(x+1)}{e^x} \right]
$$
so, for any value of $x \in(-1,0]$ (can you see from where this domain come  ?) we have two real values of $y$ and this is not a function.  We can have a function if we take only one value for the square root as:
$$
y=\sqrt{\ln\left[\frac{1-\ln(x+1)}{e^x} \right]}
$$
Second: the implicit derivative gives that:
$$
\frac{1}{x+1}+e^{x+y^2}\left(1+2y\frac{dx}{dy} \right)=0
$$
so:
$$
y'=\frac{dx}{dy}=\frac{-1}{(x+1)2ye^{x+y^2}}-\frac{1}{2y}
$$
and:
$$
\lim_{x\to0^-}y'=\infty
$$
so, at $(0,0) $ the tanget to the curve is the $y$ axis.
