Can y' be squared? $\frac{dy}{dx} ye^{xy}=3x$

Question

So the question I am trying to solve is: find $\frac{dy}{dx}$ if $ye^{xy}=3x$

I have tried:

$\frac{d}{dx} ye^{xy}=\frac{d}{dx}3x$, factor out constant, and add y':

$yy'\frac{d}{dx} e^{xy}=3$, derivative of $e^u$ is $e^u$, use chain rule on u=(xy)

$yy' e^{xy}\frac{d}{dx}(xy)=3$, Using the product rule I should be left with:

$yy' e^{xy}(xy'+(1)y)=3$, expand and factor:

$y^2(y')^2x$ $ e^{xy}=3$

$(y')^2$$=\frac{3}{ xe^{xy}y^2}$

$dy/dx$$=\sqrt\frac{3}{ xe^{xy}y^2}$

But that doesn't seem right. Any help is appreciated. Thank you.

Answer 1

$$ ye^{xy}=3x$$ Take the derivative of both sides. Note the chain rule and product rule used. $$y'e^{xy} + ye^{xy}*(y+xy')= 3$$ Factor out $e^{xy}$ $$e^{xy} (y'+y(y+xy')) = 3$$ $$e^{xy} (y'+y^2+xyy') = 3$$ $$y'+y^2+xyy'=\frac{3}{e^{xy}}$$Factor out $y'$ $$y'(1+xy)+y^2 = \frac{3}{e^{xy}}$$ $$y'(1+xy) = \frac{3}{e^{xy}} - y^2$$ $$y' = \frac{3}{e^{xy}(1+xy)} - \frac{y^2}{1+xy}$$