<s>[INST] Calculus I: Find $y'$ equation: $x^4y+\sqrt{xy}=5$ I need to find $y'$ from the equation $x^4y+\sqrt(xy)=5$
Wolffram and Mathway(pro) are giving me way different answers and I can not follow what they are doing and where I am going wrong.

My attempt:  
$$x^4y+\sqrt{xy}=5$$
$$x^4y+(xy)^{1/2} = 5$$
$$x^4y'+y(4x^3)+\frac{1}{2}(xy)^\frac{-1}{2}(xy'+y)=0$$
$$y'(x^4+\frac{1}{2}(xy)^{1/2}(x+y))=-4x^3y$$
$$y' = -4x^3y / (x^4+1/2(xy)^{1/2}(x+y))$$
 [/INST] Use multiplication rule:
$$(x^4y)'=4x^3y+x^4y'$$
$$(\sqrt{xy})'=\frac{y}{2\sqrt{xy}}+\frac{xy'}{2\sqrt{xy}}$$
$$5'=0$$
We get
$$4x^3y+x^4y'+\frac{y}{2\sqrt{xy}}+\frac{xy'}{2\sqrt{xy}}=0$$
$$x^4y'+\frac{xy'}{2\sqrt{xy}}=-4x^3y-\frac{y}{2\sqrt{xy}}$$
$$y'\bigg(x^4+\frac{x}{2\sqrt{xy}}\bigg)=-4x^3y-\frac{y}{2\sqrt{xy}}$$
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